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: Wrong Answer)


/* 実行時間1s制限の場合10^6 1,000,000 : 余裕をもって間に合う10^7
  10,000,000 : おそらく間に合う10^8 100,000,000 : 非常にシンプルな処理
  でないと厳しい
 */
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define repi(i, a, b) for(int i = int(a); i < int(b); ++i)
#include <bits/stdc++.h>
#define M_PI 3.14159265358979323846
using namespace std;

typedef long long ll; // long longはデータモデル(LLP64, LP64, ...)によらず64bit.
ll gcd(ll a, ll b) {
    return b ? gcd(b, a % b) : a;
} // Greatest Common Divisor ユークリッドの互除法（aが長辺）
vector<pair<ll, int>> factorize(ll n) { // a,bの公約数 = GCD(a, b)の約数
    vector<pair<ll, int>> res;
    for(ll i = 2; i * i <= n; i++) { // √nまで探せばよい
        if(n % i)
            continue;
        res.emplace_back(i, 0); // 割り切れたら
        while(n % i == 0) {
            n /= i;
            res.back().second++;
        }
    }
    if(n != 1)
        res.emplace_back(n, 1); // この時点で1でなかったら、残りは素数.
    return res;
}
ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
} // Least Commont Multiple オーバーフローしないように先に割る
ll ceil(ll a, ll b) { return (a + b - 1) / b; }

const ll INF = LONG_MAX;
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return 1;
    }
    return 0;
}

// https://qiita.com/drken/items/4a7869c5e304883f539b#3-3-dfs-%E3%81%AE%E5%86%8D%E5%B8%B0%E9%96%A2%E6%95%B0%E3%82%92%E7%94%A8%E3%81%84%E3%81%9F%E5%AE%9F%E8%A3%85
/*
using Graph = vector<vector<int>>;
vector<bool> seen;
// 初期化方法 seen.assign(N, false); // 全頂点を「未訪問」に初期化
void dfs(const Graph &g, int v) {
    seen[v] = true;
    for(auto next_v : g[v]) {
        if(seen[next_v])
            continue;
        dfs(g, next_v);
    }
}
*/
// ***
// 01 - BSF
// ***
// ex.
//    Edge edge[9];
//    vector<int> dist(9);
//    addEdge(true, edge, 0, 1, 0);
//    addEdge(true, edge, 0, 7, 1);
//    ...
//    int src = 0;  // source node
//    zeroOneBFS(edge, dist ,src);

const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};

int main() {

#ifdef LOCAL
    // 以降 cin の入力元が 'xxx.txt' になる
    std::ifstream in("code.txt");
    std::cin.rdbuf(in.rdbuf());
#endif
    int n;
    cin >> n;
    vector<vector<int>> e(n, vector<int>(n));
    rep(i, n) {
        rep(j, n) { cin >> e[i][j]; }
    }

    rep(k, n) {
        rep(i, n) {
            rep(j, n) {
                if(i == j || j == k || k == i)
                    continue;
                if(e[i][k] + e[k][j] < e[i][j]) {
                    cout << -1;
                    return 0;
                }
            }
        }
    }

    rep(k, n) {
        rep(i, n) {
            rep(j, n) {
                if(i == j || j == k || k == i)
                    continue;
                if(e[i][k] + e[k][j] == e[i][j]) {
                    e[i][j] = 0xFFFFFF;
                    e[j][i] = 0xFFFFFF;
                }
            }
        }
    }

    int ans = 0;
    rep(i, n) {
        rep(j, n) { if(e[i][j]!=0xFFFFFF)ans += e[i][j]; }
    }

    cout << ans / 2;

    return 0;
}

/*
  - 最大値
    long long 9,223,372,036,854,775,807 < 9.223.. * 10^18
  - 1e9 : 10^9
  - container一般
    - 合計
      accumulate(a.begin(), a.end(), 0LL)  0LLはオーバーフロー用.
  数値は合計に履かせる下駄.
  - vector
    vector<int> A(N, 0); // 0で初期化(省略可)
    vector<vector<int>> vec(n_rows, vector<int>(n_cols, value));  //
  2次元配列初期化
    - loop
      for  (int i = 0; i < A.size(); i++) {}
      for  (int elem : A) {}
    - sort
      std::sort(v.begin(), v.end());  // 昇順
      std::sort(v.begin(), v.end(), std::greater<int>()); //降順
      - vector<vector<int>> cnt(h, vector<int>(w))
    - sort(struct)
        struct st_t {
            string name;
            int p;
            bool operator<(const st_t& another) const {
                if (name == another.name)
                {
                    return p > another.p;
                }
                return name < another.name;
            }
        };
        vector<st_t> st(n);
        sort(st.begin(), st.end());
  - pair
    - pairのソートの定義 : firstが優先的に比較。同じだったらsecond。
    - pair<pair<string,int>,int> p[110];
      std::sort(p,p+a);
        こうすると、first.first -> first.second -> secondの順にソートされる
    - sort
      - sort(a.begin(), cb.end(), greater<pair<int, int>>());
  - map
    - for (auto x : map) {}  // x.first:key  x.second:value.
  - priority_queue<int> q;
    - q.top()に最大値が入っている
  - string
    - loop
      for (char& c : s) {}
 */
/*
    // エラトステネスのふるい
    int MAX = 101010;
    vector<int> is_prime(MAX, 1);
    is_prime[0] = 0, is_prime[1] = 0;
    for (int i = 2; i < MAX; ++i) {
        if (!is_prime[i]) continue;
        for (int j = i*2; j < MAX; j += i) is_prime[j] = 0;
    }
*/

#if 0
struct node {
    int to, weight;
};
using Edge = vector<node>;
void addEdge(bool directed, Edge *edges, int u, int v, int wt) {
    edges[u].push_back({v, wt});
    if(!directed) {
        edges[v].push_back({u, wt});
    }
}

int pre_node[100000];
void zeroOneBFS(Edge *edges, vector<int> &dist, int src) {
    // Initialize distances from given source
    for(int i = 0; i < dist.size(); i++)
        dist[i] = INT_MAX;
    // double ende queue to do BFS.
    deque<int> Q;
    dist[src] = 0;
    Q.push_back(src);

    while(!Q.empty()) {
        int v = Q.front();
        Q.pop_front();

        for(int i = 0; i < edges[v].size(); i++) {
            // checking for the optimal distance
            if(dist[edges[v][i].to] > dist[v] + edges[v][i].weight) {
                dist[edges[v][i].to] = dist[v] + edges[v][i].weight;
                pre_node[edges[v][i].to] = v;

                // Put 0 weight edges to front and 1 weight
                // edges to back so that vertices are processed
                // in increasing order of weights.
                if(edges[v][i].weight == 0)
                    Q.push_front(edges[v][i].to);
                else
                    Q.push_back(edges[v][i].to);
            }
        }
    }

    // printing the shortest distances
    for(int i = 0; i < dist.size(); i++)
        cout << dist[i] << " ";
}
#endif