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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<cstdint>
using namespace std;

const int MAX = 300;
const int INF = 1073741823;
int main() {
	int N;
	static int A[MAX][MAX], load[MAX][MAX];
	cin >> N;
	for (int i = 0;i < N;i ++) {
		for (int j = 0;j < N;j ++) {
			cin >> A[i][j];
			load[i][j] = A[i][j];
		}
	}
	for (int i = 0;i < N;i ++) {
		for (int j = 0;j < N;j ++) {
			for (int k = 0;k < N;k ++) {
				load[i][j] = min(load[i][j], load[i][k] + load[k][j]); // ワーシャルフロイド法
			}
		}
	}
	int64_t dist = 0;
	bool add;
	for (int i = 0;i < N;i ++) {
		for (int j = 0;j < N;j ++) {
			if (i == j) continue;
			if (A[i][j] > load[i][j]) { // この時点でこんな経路が無いことが分かるよ
				cout << -1;
				return 0;
			}
			add = true;
			for (int k = 0;k < N;k ++) {
				if (i == k || j == k) continue;
				if (load[i][j] == load[i][k] + load[k][j]) add = false; // 要らない辺
			}
			if (add) dist += load[i][j];
		}
	}
	cout << dist / 2; // 同じ辺を行き帰りで2回数えているので
	return 0;
}
/*
31536000のコメント解説欄
ここテンプレで用意してるから、A問題とかだとこの先空欄の危険あり

まず、先に道を置いておこう
そして、最短経路問題として、つまりWarshall-Floyd法を用いて最短経路を確定させよう
後は元のと比較して、うっかり元経路より短くしてたら終了、-1
そうでなければ、必要な辺だけ数えていこう

次に、必要な辺とは何か考えよう
もしその辺が無くなるとコストが増えるなら、それは必要な辺だよね
逆にその辺が無くなってもコスト増えないなら、まぁそれはうん、いらねぇ
*/