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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<n;++i)
#define exrep(i, a, b) for(ll i = a; i < b; i++)
#define out(x) cout << x << endl
#define EPS (1e-7)
#define gearup ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<pair<int,int> > vpii;
typedef vector<vector<int> > vvi;
typedef vector<vector<char> > vvc;
typedef vector<vector<bool> > vvb;
typedef vector<vector<double> > vvd;
typedef vector<vector<string> > vvs;
typedef vector<ll> vl;
typedef vector<vector<ll> > vvl;
typedef vector<vector<vector<ll> > > vvvl;
ll MOD = 1000000007;
const long long L_INF = 1LL << 60;
const int INF = 2147483647; // 2^31-1
const double PI = acos(-1);
//cout<<fixed<<setprecision(10);

template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b;return true;}return false;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b;return true;}return false;}
template<class T> void debug(T v){rep(i,v.size()) cout<<v[i]<<" ";cout<<endl;}
const ll dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
//以降 cin の入力元が 'input.txt' になる
//std::ifstream in("input.txt");
//std::cin.rdbuf(in.rdbuf());

struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
    vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)

    UnionFind(int N) : par(N),siz(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++){
             par[i] = i;
             siz[i] = 1;
        }
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        // merge technique（小を大にくっつける）
        if (siz[rx] < siz[ry]) swap(rx, ry);
        siz[rx] += siz[ry];
        par[ry] = rx;
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }

    ll size(ll x){
        return siz[root(x)];
    }
};

signed main()
{
    gearup;
    int n; cin >> n;
    vvl a(n,vl(n));
    rep(i,n)rep(j,n)cin>>a[i][j];
    auto dp = a;
    vvb b(n,vb(n,true));
    //ワーシャルフロイド法
    for(int k=0;k<n;k++){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(dp[i][j] > (dp[i][k] + dp[k][j]) ){ //矛盾発見
                    out(-1);
                    return 0;
                }
                if(dp[i][j] == dp[i][k] + dp[k][j] && dp[i][k] > 0 && dp[k][j] > 0)
                    b[i][j] = false; //迂回しても同じ距離なので迂回した方を採用
            }
        }
    }
    ll res = 0;
    rep(i,n)rep(j,n)if(b[i][j])res += dp[i][j];
    out(res/2);
}