Problem:
Score : 500 points

Problem Statement

In Takahashi Kingdom, which once existed, there are N cities, and some pairs of cities are connected bidirectionally by roads.
The following are known about the road network:

People traveled between cities only through roads. It was possible to reach any city from any other city, via intermediate cities if necessary.

Different roads may have had different lengths, but all the lengths were positive integers.

Snuke the archeologist found a table with N rows and N columns, A, in the ruin of Takahashi Kingdom.
He thought that it represented the shortest distances between the cities along the roads in the kingdom.

Determine whether there exists a road network such that for each u and v, the integer A_{u, v} at the u-th row and v-th column of A is equal to the length of the shortest path from City u to City v.
If such a network exist, find the shortest possible total length of the roads.

Constraints

1 \leq N \leq 300

If i ≠ j, 1 \leq A_{i, j} = A_{j, i} \leq 10^9.

A_{i, i} = 0

Inputs

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
...
A_{N, 1} A_{N, 2} ... A_{N, N}

Outputs

If there exists no network that satisfies the condition, print -1.
If it exists, print the shortest possible total length of the roads.

Sample Input 1

3
0 1 3
1 0 2
3 2 0

Sample Output 1

3

The network below satisfies the condition:

City 1 and City 2 is connected by a road of length 1.

City 2 and City 3 is connected by a road of length 2.

City 3 and City 1 is not connected by a road.

Sample Input 2

3
0 1 3
1 0 1
3 1 0

Sample Output 2

-1

As there is a path of length 1 from City 1 to City 2 and City 2 to City 3, there is a path of length 2 from City 1 to City 3.
However, according to the table, the shortest distance between City 1 and City 3 must be 3.

Thus, we conclude that there exists no network that satisfies the condition.

Sample Input 3

5
0 21 18 11 28
21 0 13 10 26
18 13 0 23 13
11 10 23 0 17
28 26 13 17 0

Sample Output 3

82

Sample Input 4

3
0 1000000000 1000000000
1000000000 0 1000000000
1000000000 1000000000 0

Sample Output 4

3000000000

Solution in PyPy3 (2.4.0) (judge verdict: Wrong Answer)

#!/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
        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()