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 <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <fstream> 
#define _USE_MATH_DEFINES
#include <math.h>
#include <unordered_set>

using namespace std;


inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }

template<class T> inline T sqr(T x) { return x*x; }

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<long long int> vll;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  FOR(i,0,n)
#define P(p) cout<<(p)<<endl;
#define VEC_2D(a,b) vector<vector<int> >(a, vector<int>(b, 0))
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define pb push_back
#define mp make_pair
#define INF (1000000000)
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define MOD 1000000007LL
#define FSP(a) cout << fixed << setprecision(a)


template<typename T> T gcd(T x, T y) {
	if (y == 0) return x;
	else return gcd(y, x%y);
}

template<typename T> T lcm(T a, T b) {
	return a / gcd(a, b) * b;
}
template<typename T> bool is_prime(T n) {
	for (int i = 2; i * i <= n; i++) {
		if (n % i == 0) return false;
	}
	return n != 1;
}
map<int, int> prime_factor(int n) {
	map<int, int> res;
	for (int i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			++res[i];
			n /= i;
		}
	}
	if (n != 1) res[n] = 1;
	return res;
}
int extgcd(int a, int b, int& x, int& y) {//
	int d = a;
	if (b != 0) {
		d = extgcd(b, a%b, y, x);
		y -= (a / b)*x;
	}
	else {
		x = 1; y = 0;
	}
	return d;
}



ll mod_pow(ll x, ll n, ll mod) {
	if (n == 0) return 1;
	ll res = mod_pow(x * x % mod, n / 2, mod);
	if (n & 1) res = res * x % mod;
	return res;
}
vector<string> split(const string &str, char delim) {
	vector<string> res;
	size_t current = 0, found;
	while ((found = str.find_first_of(delim, current)) != string::npos) {
		res.push_back(string(str, current, found - current));
		current = found + 1;
	}
	res.push_back(string(str, current, str.size() - current));
	return res;
}

bool is_kadomatsu(int a, int b, int c) {
	if (a == b || a == c || b == c)return false;
	if (a > b && c > b) return true;
	if (a < b && c < b)return true;
	return false;
}


struct UF {
	int n;
	vi d;
	UF() {}
	UF(int n) :n(n), d(n, -1) {}
	int root(int v) {
		if (d[v] < 0) return v;
		return d[v] = root(d[v]);
	}
	bool same(int a, int b) {
		return root(a) == root(b);
	}
	bool unite(int x, int y) {
		x = root(x); y = root(y);
		if (x == y) return false;
		if (size(x) < size(y)) swap(x, y);
		d[x] += d[y];
		d[y] = x;
		return true;
	}
	int size(int v) { return -d[root(v)]; }
};



vector<int> divisor(int n) {
	if (n == 1) return{};
	vi res;
	for (int i = 1; i*i <= n; i++) {
		if (n%i == 0) {
			res.emplace_back(i);
			if (i != 1 && i != n / i)res.emplace_back(n / i);
		}
	}
	return res;
}

struct Bellmanford {
int n;
struct edge {
	int from, to, cost;
};
vector<edge> E;
vi d;
Bellmanford(int n) :n(n), d(n) {
	E.resize(n);
}
void add_edge(int x, int y, int cost) {
	edge e;
	e.from = x; e.to = y; e.cost = cost;
	E.push_back(e);
}
void shortest_path(int s) {
	rep(i, n)d[i] = INF;
	d[s] = 0;
	while (true) {
		bool update = false;
		for (auto e : E) {
			if (d[e.from] != INF && d[e.to] > d[e.from] + e.cost) {
				d[e.to] = d[e.from] + e.cost;
				update = true;
			}
		}
		if (!update) break;
	}
}
};
struct Dijkstra {
	int n;
	struct edge {
		int to; ll cost;
	};
	vector<vector<edge>> G;
	vll d;
	Dijkstra(int n) :n(n), d(n) {
		G.resize(n);
	}
	void add_edge(int x, int y, ll cost) {
		edge e;
		e.to = y; e.cost = cost;
		G[x].push_back(e);
	}
	void shortest_path(int s) {
		rep(i, n)d[i] = 100000000000000000;
		d[s] = 0;
		priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> que;
		que.push(make_pair(0, s));

		while (!que.empty()) {
			pii p = que.top(); que.pop();
			int v = p.second;
			if (d[v] < p.first) continue;
			for (auto e : G[v]) {
				if (d[e.to] > d[v] + e.cost) {
					d[e.to] = d[v] + e.cost;
					que.push(make_pair(d[e.to], e.to));
				}
			}
		}
	}
};

struct Segmenttree {
	int n;
	vector<pair<ll, int>> dat;
	Segmenttree() {}
	void init(ll n_) {
		n = 1;
		while (n < n_) n *= 2;
		dat.resize(2 * n - 1);
		rep(i, 2 * n - 1)dat[i] = pair<ll, int>(-INF, -INF);
	}
	void update(int idx, ll val) {
		idx += n - 1;
		dat[idx] = make_pair(val, -(idx - n + 1));
		while (idx > 0) {
			idx = (idx - 1) / 2;
			dat[idx] = max(dat[idx * 2 + 1], dat[idx * 2 + 2]);
		}
	}
	pair<ll, int> query(int a, int b) {
		return query_seg(a, b, 0, 0, n);
	}
	pair<ll, int> query_seg(int a, int b, int k, int l, int r) {
		if (r <= a || b <= l) return pair<ll, int>(-INF, -INF);
		if (a <= l && r <= b)return dat[k];
		else {
			return max(query_seg(a, b, k * 2 + 1, l, (l + r) / 2), query_seg(a, b, k * 2 + 2, (l + r) / 2, r));
		}
	}
};
//-------------------------------------------------------------



int main() {
	
	int N;
	cin >> N;
	int m[301][301];
	rep(i, N)rep(j, N) cin >> m[i][j];
	int d[301][301];
	rep(i, 301)rep(j, 301) {
		d[i][j] = m[i][j];
	}
	rep(k, N) {
		rep(i, N) {
			rep(j, N) {
				if (d[i][j] > d[i][k] +d[k][j]) {
					d[i][j] = d[i][k] + d[k][j];
					if (d[i][j] < m[i][j]) {
						P(-1);
						return 0;
					}
				}
			}
		}
	}
	ll ans = 0;
	rep(i, N) {
		rep(j, N) {
			if (i == j) continue;
			bool use = true;
			rep(w, N) {
				if (i == w || j == w) continue;
				if (d[i][j] == d[i][w] + d[w][j]) use = false;
			}
			if (use)ans += (ll)d[i][j];
		}
	}
	P(ans/2);
	return 0;

}