#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 (int j = 1; j <= n; j++) {
      if (dis1[i][j] != dis2[i][j])
        ans++;
    }
  }
  cout << ans << endl;
  return 0;
}

以上程序的输入数据满足 $1 \leq n \leq 100,1 \leq m \leq \dfrac{n(n-1)}{2}$，且只保证不存在重边，即不存在 $(u_i,v_i)=(u_j,v_j)(i \neq j)$，边权 $w_i \in [1,10^6]$ 。如果 $u$ 到 $v$ 不可达，则认为距离为 $\texttt{INF}$ 。完成下面的判断题和单选题。

第5题的输入数据：

5 8
3 2 2
2 4 2
1 4 3
3 1 2
4 3 3
5 2 3
1 5 1
1 2 2