这个程序的时间复杂度和m无关。（ ）

#include <cstdio>
#include <cstring>
using namespace std;

int n, m;
int f[101][101];
int F[101][101];

int main() {
    scanf("%d%d", &n, &m); // n的值在1到100之间
    memset(f, -1, sizeof(f));
    for(int i = 1; i <= m; i++) {
        int u, v, w; // w的值在0到10000之间
        scanf("%d%d%d", &u, &v, &w);
        f[u][v] = f[v][u] = w;
    }

    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++) {
                if(f[i][k] == -1 && f[k][j] == -1)
                    continue;
                if(f[i][j] == -1 || f[i][j] > f[i][k] + f[k][j])
                    f[i][j] = f[i][k] + f[k][j];
            }
    int ans = 2147483647;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            for(int x = 1; x <= n; x++)
                for(int y = 1; y <= n; y++)
                    F[x][y] = f[x][y];
            F[i][j] = F[j][i] = 0;

            for(int x = 1; x <= n; x++)
                for(int y = 1; y <= n; y++)
                    if(F[x][y] == -1 || F[x][y] > F[x][i] + F[i][y])
                        F[x][y] = F[x][i] + F[i][y];

            int res = 0;
            for(int x = 1; x <= n; x++)
                for(int y = 1; y <= n; y++)
                    res += F[x][y];
            ans = min(res, ans);
        }
    }
    printf("%d\n", ans);
    return 0;
}