A61546 | 阅读程序：#include <bits/stdc++.h>

阅读程序：

#include <bits/stdc++.h>
using namespace std;
#define N 105
#define INF 1e9
int dis1[N][N], dis2[N][N];
int mp[N][N], n, m;
void fun1(int dis[N][N]) {
    static bool vis[N];
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            dis[i][j] = mp[i][j];
        } 
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) vis[j] = 0;
          for (int k = 1; k <= n; k++) {
            int now = 0;
            for (int j = 1; j <= n; j++) {
                if (!vis[j] && (!now || dis[i][now] > dis[i][j])) now = j;
    }
    vis[now] = 1;
    for (int j = 1; j <= n; j++) {
        if(!vis[j]&&dis[i][j] > dis[i][now]+mp[now][j]){
           dis[i][j] = dis[i][now] + mp[now][j];
                }
            }
        }
    }
}
void fun2(int dis[N][N]) {
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            dis[i][j] = mp[i][j];
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
          for (int k = 1; k <= n; k++) {
            dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
            }
        }
    }
}
int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (i == j) mp[i][j] = 0;
            else mp[i][j] = INF;
        }
    }
for (int i = 1; i <= m; i++) {
  int u, v, w;
  cin >> u >> v >> w;
  mp[u][v] = w;
}
fun1(dis1);
fun2(dis2);
int ans = 0;
for (int i = 1; i <= n; i++) {
    for(intj=1;j<=n;j++){
        if (dis1[i][j] != dis2[i][j])
            ans++;
        }
    }
cout << ans << endl;
return 0;
}

以下程序的输入数据满足 𝟏 ≤ 𝒏 ≤ 𝟏𝟎𝟎, 𝟏 ≤ 𝒎 ≤ 𝒏(𝒏−𝟏) ，且只保证不存在 𝟐重边，即不存在 (𝒖𝒊, 𝒗𝒊) = (𝒖𝒋, 𝒗𝒋) (𝒊 ≠ 𝒋)，边权 𝒘𝒊 ∈ [𝟏, 𝟏𝟎𝟔] 。如果 u 到 v不可达，则认为距离为 INF 。完成下面的判断题和单选题: