Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3

import sys
# import time
# import math
# import numpy as np
# import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random                                # random, uniform, randint, randrange, shuffle, sample
# import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
# from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate             # accumulate(iter[, f])
from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from fractions import gcd                    # for Python 3.4 (previous contest @AtCoder)



def main():
    mod = 1000000007                # 10^9+7
    inf = float('inf')              # sys.float_info.max = 1.79...e+308
    # inf = 2 ** 64 - 1             # (for fast JIT compile in PyPy) 1.84...e+19
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())


    # class PQueueMin:
    #     def __init__(self):
    #         self.pq = []    # 第一要素に cost、第二要素に ind が来る様にする
        
    #     def push(self, ind, cost):
    #         heappush(self.pq, [cost, ind])
        
    #     def pop(self):
    #         cost, ind = heappop(self.pq)
    #         return [ind, cost]

    #     def is_empty(self):
    #         return len(self.pq)==0



    # def dijkstra(adj_with_weight, start, goal):
    #     '''
    #     Dijkstra 法 O((E+V)lgV) (頂点を回るときに最小のものを取り出していくのに VlgV, cost の更新で ElgV)
    #     負辺を含まぬ重みつきグラフについて、ある頂点から他の頂点までの最短コストを貪欲に計算する
    #     '''
    #     V = len(adj_with_weight)
    #     cost = [inf] * V
    #     cost[start] = 0
    #     fixed = [False] * V
    #     pq = PQueueMin()
    #     # 最初は始点が必ずコスト最小。pq に突っ込む
    #     pq.push(start, cost[start])
    #     while not pq.is_empty():
    #         # 現段階で最短のコストとなっている頂点を選択
    #         u, cost_of_u = pq.pop()
    #         fixed[u] = True
    #         # 枝刈り
    #         if u == goal:
    #             return cost[goal]
    #         for v, weight_of_uv in adj_with_weight[u]:
    #             if not fixed[v] and cost[v] > cost_of_u + weight_of_uv:
    #                 cost[v] = cost_of_u + weight_of_uv
    #                 pq.push(v, cost[v])
    #     return inf


    class UnionFindTree:
        def __init__(self, num_of_elm):
            '''
            0 ... num_of_elem - 1 まで数字で表される (0-index) グループを管理する union find tree を作成する (O(n))
            '''
            self.n = num_of_elm
            self.table = [i for i in range(self.n)]    # table[ind] は ind の親の index を表す。ind == table[ind] の時そのグループの root である。
            self.rank = [0] * self.n    # root となる ind について rank[ind] はその木の深さを表す。それ以外の ind については意味を持たない。
            self.group_size = [1] * self.n    # root となる ind について group_size[ind] はそのグループに属するメンバの個数を表す。それ以外の ind については意味を持たない。
        
        def _find_set(self, x):
            '''
            x の属するグループ番号を O(α(n)) で求める
            '''
            parent = self.table[x]
            if x == parent:
                return x
            else:
                root = self._find_set(parent)
                # 経路圧縮
                self.table[x] = root
                return root
        
        def is_same(self, x, y):
            '''
            x と y が同じグループに属するか O(α(n)) で判定する
            '''
            return self._find_set(x) == self._find_set(y)
    
        def union(self, x, y):
            '''
            x と y の属するグループを O(α(n)) で統合する
            '''
            shallow_root = self._find_set(x)
            deep_root = self._find_set(y)
            if self.rank[shallow_root] > self.rank[deep_root]:
                shallow_root, deep_root = deep_root, shallow_root
            # そもそも同一グループだった時
            if shallow_root == deep_root:
                return False
            # グループが異なるので union
            else:
                self.table[shallow_root] = deep_root
                self.group_size[deep_root] += self.group_size[shallow_root]
                # 深さが等しかったときはつけ加えられた側の rank をインクリメントする
                if self.rank[shallow_root] == self.rank[deep_root]:
                    self.rank[deep_root] += 1
                return True
    
        def akin_num(self, x):
            '''
            x の属するグループのサイズを O(1) で計算する
            '''
            x_root = self._find_set(x)
            return self.group_size[x_root]
    
        def print_group_id(self):
            print([self._find_set(x) for x in self.table])


    # def update_matrix_joint(fixed_dist_matrix, i, j, w):
    #     """
    #     (i, j がすでに連結な時)
    #     i, j 間に新たに w の距離の道をはる。距離行列を更新する。
    #     """
    #     fixed_dist_matrix[i][j] = w
    #     fixed_dist_matrix[j][i] = w
    #     for k in range(n):
    #         if k != i:
    #             new_dist = min(fixed_dist_matrix[k][i], fixed_dist_matrix[k][j] + w)
    #             fixed_dist_matrix[k][i] = new_dist
    #             fixed_dist_matrix[i][k] = new_dist
    #         if k != j:
    #             new_dist = min(fixed_dist_matrix[k][j], fixed_dist_matrix[k][i] + w)
    #             fixed_dist_matrix[k][j] = new_dist
    #             fixed_dist_matrix[j][k] = new_dist
    
    def update_matrix_disjoint(fixed_dist_matrix, i, j, w, i_group, j_group):
        """
        i, j 間に新たに w の距離の道をはる。距離行列を更新する。        
        """
        fixed_dist_matrix[i][j] = w
        fixed_dist_matrix[j][i] = w
        for s, t in product(i_group, j_group):
            if s != i or t != j:
                new_dist = min(fixed_dist_matrix[s][t], fixed_dist_matrix[s][i] + w + fixed_dist_matrix[j][t])
                # print(f"{s} {t} {new_dist}")
                fixed_dist_matrix[s][t] = new_dist
                fixed_dist_matrix[t][s] = new_dist


    
    def struct_street(edges):
        n = len(adj_mat)
        total_length = 0
        # adj_list = [[] for _ in range(n)]
        
        uf = UnionFindTree(n)
        fixed_dist_matrix = [[inf] * n for _ in range(n)]
        for i in range(n):
            fixed_dist_matrix[i][i] = 0

        for (i, j), w in edges:
            # dist = dijkstra(adj_list, start=i, goal=j)
            # print(fixed_dist_matrix)
            dist = fixed_dist_matrix[i][j]
            if dist > w:
                # 辺をはる
                # adj_list[i].append((j, w))
                # adj_list[j].append((i, w))
                # if uf.is_same(i, j):
                #     update_matrix_joint(fixed_dist_matrix, i, j, w)
                # else:
                i_group = []
                j_group = []
                for ind in range(n):
                    if uf.is_same(i, ind):
                        i_group.append(ind)
                    if uf.is_same(j, ind):
                        j_group.append(ind)
                update_matrix_disjoint(fixed_dist_matrix, i, j, w, i_group, j_group)
                uf.union(i, j)
                total_length += w
            elif dist == w:
                # スルー
                continue
            else:
                # 構築不可能
                return -1
        return total_length

    


    n = ii()
    adj_mat = [lmi() for _ in range(n)]

    edges = []
    for i in range(n):
        for j in range(i+1, n):
            tmp = [(i, j), adj_mat[i][j]]
            edges.append(tmp)
    edges.sort(key=itemgetter(1))

    print(struct_street(edges))



if __name__ == "__main__":
    main()


Fixed submission:
#!/usr/bin/env python3

import sys
# import time
# import math
# import numpy as np
# import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random                                # random, uniform, randint, randrange, shuffle, sample
# import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate             # accumulate(iter[, f])
# from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from fractions import gcd                    # for Python 3.4 (previous contest @AtCoder)



def main():
    mod = 1000000007                # 10^9+7
    inf = float('inf')              # sys.float_info.max = 1.79...e+308
    # inf = 2 ** 64 - 1             # (for fast JIT compile in PyPy) 1.84...e+19
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())


    def warshall_floyd(adj_mat_with_weight):
        '''
        Warshall-Floyd 法 (O(V^3), 1 sec だと v~500 とかが限界)
        全点対間の最短経路を三次元 DP で求める
            - 定式化
            - dp[k+1][i][j] を 0...k, i, j なる頂点を使用できる時の i~j 最短距離とする (k+1=0 の時は e(i,j))
            - dp[k+1][i][j] = min(dp[k][i][j], dp[k][i][k+1]+dp[k][k+1][j])
            - 最後に必要なのは dp[V][i][j] 。上の漸化式をみるに 2 次元 array があれば使いまわしてメモリ節約できる。
        Args:
            adj_mat_with_weight (list): 重みを記録した隣接行列 (非接続頂点は inf)
        Returns:
            dp (list): dp[i][j] を見れば全頂点を使用可能な時の i j 最短距離がわかる。最後に dp[i][i] が負になっていたら負サイクルが存在するということ
        '''
        V = len(adj_mat_with_weight)
        # テーブル初期化
        dp = deepcopy(adj_mat_with_weight)
        for i in range(V):
            dp[i][i] = 0
        # 漸化式を解く
        for k in range(1, V+1):
            for i in range(V):
                for j in range(V):
                    # new 2d-table # old 2d-table
                    dp[i][j] = min(dp[i][j], dp[i][k-1]+dp[k-1][j])
        return dp

    def check_possible_structure(dp, adj_mat):
        n = len(adj_mat)
        for i in range(n):
            for j in range(i+1, n):
                if dp[i][j] < adj_mat[i][j]:
                    return False
        return True

    def calc_minimum_street(dp, adj_mat):
        n = len(adj_mat)
        total_dist = 0
        for i in range(n):
            for j in range(i+1, n):
                total_dist += adj_mat[i][j]
        # 削除していく
        for i in range(n):
            for j in range(i+1, n):
                # (i, j) の距離について
                dist = adj_mat[i][j]
                for k in range(n):
                    if k != i and k != j and dp[i][k] + dp[k][j] == dist:
                        total_dist -= dist
                        break
        return total_dist
    
    
    n = ii()
    adj_mat = [lmi() for _ in range(n)]
    dp = warshall_floyd(adj_mat)

    if not check_possible_structure(dp, adj_mat):
        print(-1)
    else:
        print(calc_minimum_street(dp, adj_mat))


if __name__ == "__main__":
    main()

Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3

import sys
# import time
# import math
# import numpy as np
# import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random                                # random, uniform, randint, randrange, shuffle, sample
# import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
# from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate             # accumulate(iter[, f])
from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from fractions import gcd                    # for Python 3.4 (previous contest @AtCoder)



def main():
    mod = 1000000007                # 10^9+7
    # inf = float('inf')              # sys.float_info.max = 1.79...e+308
    inf = 2 ** 64 - 1             # (for fast JIT compile in PyPy) 1.84...e+19
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())


    class PQueueMin:
        def __init__(self):
            self.pq = []    # 第一要素に cost、第二要素に ind が来る様にする
        
        def push(self, ind, cost):
            heappush(self.pq, [cost, ind])
        
        def pop(self):
            cost, ind = heappop(self.pq)
            return [ind, cost]

        def is_empty(self):
            return len(self.pq)==0



    def dijkstra(adj_with_weight, start, goal):
        '''
        Dijkstra 法 O((E+V)lgV) (頂点を回るときに最小のものを取り出していくのに VlgV, cost の更新で ElgV)
        負辺を含まぬ重みつきグラフについて、ある頂点から他の頂点までの最短コストを貪欲に計算する
        '''
        V = len(adj_with_weight)
        cost = [inf] * V
        cost[start] = 0
        fixed = [False] * V
        pq = PQueueMin()
        # 最初は始点が必ずコスト最小。pq に突っ込む
        pq.push(start, cost[start])
        while not pq.is_empty():
            # 現段階で最短のコストとなっている頂点を選択
            u, cost_of_u = pq.pop()
            fixed[u] = True
            # 枝刈り
            if u == goal:
                return cost[goal]
            
            for v, weight_of_uv in adj_with_weight[u]:
                if not fixed[v] and cost[v] > cost_of_u + weight_of_uv:
                    cost[v] = cost_of_u + weight_of_uv
                    pq.push(v, cost[v])
        return inf

    
    def struct_street(edges):
        n = len(adj_mat)
        total_length = 0
        adj_list = [[] for _ in range(n)]

        for (i, j), w in edges:
            dist = dijkstra(adj_list, start=i, goal=j)
            if dist > w:
                # 辺をはる
                adj_list[i].append((j, w))
                adj_list[j].append((i, w))
                total_length += w
            elif dist == w:
                # スルー
                continue
            else:
                # 構築不可能
                return -1
        return total_length

    


    n = ii()
    adj_mat = [lmi() for _ in range(n)]

    edges = []
    for i in range(n):
        for j in range(i+1, n):
            tmp = [(i, j), adj_mat[i][j]]
            edges.append(tmp)
    edges.sort(key=itemgetter(1))

    print(struct_street(edges))



if __name__ == "__main__":
    main()


Fixed submission:
#!/usr/bin/env python3

import sys
# import time
# import math
# import numpy as np
# import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random                                # random, uniform, randint, randrange, shuffle, sample
# import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate             # accumulate(iter[, f])
# from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from fractions import gcd                    # for Python 3.4 (previous contest @AtCoder)



def main():
    mod = 1000000007                # 10^9+7
    inf = float('inf')              # sys.float_info.max = 1.79...e+308
    # inf = 2 ** 64 - 1             # (for fast JIT compile in PyPy) 1.84...e+19
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())


    def warshall_floyd(adj_mat_with_weight):
        '''
        Warshall-Floyd 法 (O(V^3), 1 sec だと v~500 とかが限界)
        全点対間の最短経路を三次元 DP で求める
            - 定式化
            - dp[k+1][i][j] を 0...k, i, j なる頂点を使用できる時の i~j 最短距離とする (k+1=0 の時は e(i,j))
            - dp[k+1][i][j] = min(dp[k][i][j], dp[k][i][k+1]+dp[k][k+1][j])
            - 最後に必要なのは dp[V][i][j] 。上の漸化式をみるに 2 次元 array があれば使いまわしてメモリ節約できる。
        Args:
            adj_mat_with_weight (list): 重みを記録した隣接行列 (非接続頂点は inf)
        Returns:
            dp (list): dp[i][j] を見れば全頂点を使用可能な時の i j 最短距離がわかる。最後に dp[i][i] が負になっていたら負サイクルが存在するということ
        '''
        V = len(adj_mat_with_weight)
        # テーブル初期化
        dp = deepcopy(adj_mat_with_weight)
        for i in range(V):
            dp[i][i] = 0
        # 漸化式を解く
        for k in range(1, V+1):
            for i in range(V):
                for j in range(V):
                    # new 2d-table # old 2d-table
                    dp[i][j] = min(dp[i][j], dp[i][k-1]+dp[k-1][j])
        return dp

    def check_possible_structure(dp, adj_mat):
        n = len(adj_mat)
        for i in range(n):
            for j in range(i+1, n):
                if dp[i][j] < adj_mat[i][j]:
                    return False
        return True

    def calc_minimum_street(dp, adj_mat):
        n = len(adj_mat)
        total_dist = 0
        for i in range(n):
            for j in range(i+1, n):
                total_dist += adj_mat[i][j]
        # 削除していく
        for i in range(n):
            for j in range(i+1, n):
                # (i, j) の距離について
                dist = adj_mat[i][j]
                for k in range(n):
                    if k != i and k != j and dp[i][k] + dp[k][j] == dist:
                        total_dist -= dist
                        break
        return total_dist
    
    
    n = ii()
    adj_mat = [lmi() for _ in range(n)]
    dp = warshall_floyd(adj_mat)

    if not check_possible_structure(dp, adj_mat):
        print(-1)
    else:
        print(calc_minimum_street(dp, adj_mat))


if __name__ == "__main__":
    main()

Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3

import sys
# import time
# import math
# import numpy as np
# import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random                                # random, uniform, randint, randrange, shuffle, sample
# import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
# from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate             # accumulate(iter[, f])
from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from fractions import gcd                    # for Python 3.4 (previous contest @AtCoder)



def main():
    mod = 1000000007                # 10^9+7
    inf = float('inf')              # sys.float_info.max = 1.79...e+308
    # inf = 2 ** 64 - 1             # (for fast JIT compile in PyPy) 1.84...e+19
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())


    class PQueueMin:
        def __init__(self):
            self.pq = []    # 第一要素に cost、第二要素に ind が来る様にする
        
        def push(self, ind, cost):
            heappush(self.pq, [cost, ind])
        
        def pop(self):
            cost, ind = heappop(self.pq)
            return [ind, cost]

        def is_empty(self):
            return len(self.pq)==0



    def dijkstra(adj_with_weight, start, goal):
        '''
        Dijkstra 法 O((E+V)lgV) (頂点を回るときに最小のものを取り出していくのに VlgV, cost の更新で ElgV)
        負辺を含まぬ重みつきグラフについて、ある頂点から他の頂点までの最短コストを貪欲に計算する
        '''
        V = len(adj_with_weight)
        cost = [float('inf')] * V
        cost[start] = 0
        fixed = [False] * V
        pq = PQueueMin()
        # 最初は始点が必ずコスト最小。pq に突っ込む
        pq.push(start, cost[start])
        while not pq.is_empty():
            # 現段階で最短のコストとなっている頂点を選択
            u, cost_of_u = pq.pop()
            fixed[u] = True
            for v, weight_of_uv in adj_with_weight[u]:
                if not fixed[v] and cost[v] > cost_of_u + weight_of_uv:
                    cost[v] = cost_of_u + weight_of_uv
                    pq.push(v, cost[v])
        return cost[goal]

    
    
    def struct_street(adj_mat):
        n = len(adj_mat)
        total_length = 0
        adj_list = [[] for _ in range(n)]
        edges = []
        for i in range(n):
            for j in range(i+1, n):
                tmp = [(i, j), adj_mat[i][j]]
                edges.append(tmp)
        edges.sort(key=itemgetter(1))

        for (i, j), w in edges:
            dist = dijkstra(adj_list, start=i, goal=j)
            if dist > w:
                # 辺をはる
                adj_list[i].append((j, w))
                adj_list[j].append((i, w))
                total_length += w
            elif dist == w:
                # スルー
                continue
            else:
                # 構築不可能
                return -1
        return total_length

    
        


    n = ii()
    adj_mat = [lmi() for _ in range(n)]
    print(struct_street(adj_mat))



if __name__ == "__main__":
    main()


Fixed submission:
#!/usr/bin/env python3

import sys
# import time
# import math
# import numpy as np
# import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random                                # random, uniform, randint, randrange, shuffle, sample
# import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate             # accumulate(iter[, f])
# from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from fractions import gcd                    # for Python 3.4 (previous contest @AtCoder)



def main():
    mod = 1000000007                # 10^9+7
    inf = float('inf')              # sys.float_info.max = 1.79...e+308
    # inf = 2 ** 64 - 1             # (for fast JIT compile in PyPy) 1.84...e+19
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())


    def warshall_floyd(adj_mat_with_weight):
        '''
        Warshall-Floyd 法 (O(V^3), 1 sec だと v~500 とかが限界)
        全点対間の最短経路を三次元 DP で求める
            - 定式化
            - dp[k+1][i][j] を 0...k, i, j なる頂点を使用できる時の i~j 最短距離とする (k+1=0 の時は e(i,j))
            - dp[k+1][i][j] = min(dp[k][i][j], dp[k][i][k+1]+dp[k][k+1][j])
            - 最後に必要なのは dp[V][i][j] 。上の漸化式をみるに 2 次元 array があれば使いまわしてメモリ節約できる。
        Args:
            adj_mat_with_weight (list): 重みを記録した隣接行列 (非接続頂点は inf)
        Returns:
            dp (list): dp[i][j] を見れば全頂点を使用可能な時の i j 最短距離がわかる。最後に dp[i][i] が負になっていたら負サイクルが存在するということ
        '''
        V = len(adj_mat_with_weight)
        # テーブル初期化
        dp = deepcopy(adj_mat_with_weight)
        for i in range(V):
            dp[i][i] = 0
        # 漸化式を解く
        for k in range(1, V+1):
            for i in range(V):
                for j in range(V):
                    # new 2d-table # old 2d-table
                    dp[i][j] = min(dp[i][j], dp[i][k-1]+dp[k-1][j])
        return dp

    def check_possible_structure(dp, adj_mat):
        n = len(adj_mat)
        for i in range(n):
            for j in range(i+1, n):
                if dp[i][j] < adj_mat[i][j]:
                    return False
        return True

    def calc_minimum_street(dp, adj_mat):
        n = len(adj_mat)
        total_dist = 0
        for i in range(n):
            for j in range(i+1, n):
                total_dist += adj_mat[i][j]
        # 削除していく
        for i in range(n):
            for j in range(i+1, n):
                # (i, j) の距離について
                dist = adj_mat[i][j]
                for k in range(n):
                    if k != i and k != j and dp[i][k] + dp[k][j] == dist:
                        total_dist -= dist
                        break
        return total_dist
    
    
    n = ii()
    adj_mat = [lmi() for _ in range(n)]
    dp = warshall_floyd(adj_mat)

    if not check_possible_structure(dp, adj_mat):
        print(-1)
    else:
        print(calc_minimum_street(dp, adj_mat))


if __name__ == "__main__":
    main()