A 最少需要多少钱使得转账后 B 收到 100 元

100 = d(A) * W1 * W2 * … Wn

要让 A的值最小 则 (W1 * W2 * … Wn) 最大

#include<iostream>
#include<cstring>

using namespace std;
const int N = 2010, INF = 0x3f3f3f3f;
int n, m, start, ed;
double g[N][N], d[N];
bool st[N];

void dj() {
    memset(d, 0x3f, sizeof d);
    d[start] = 1;
    for (int i = 0; i < n; i++) {
        int t = -1;
        for (int j = 1; j <= n; j++) {
            if (!st[j] && (t == -1 || d[j] > d[t]))
                t = j;
        }

        st[t] = true;

        for (int j = 1; j <= n; j++) {
            d[j] = max(d[j], d[t] * g[t][j]);
        }
    }
}

int main() {
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        double z = (100.0 - c) / 100;
        g[a][b] = g[b][a] = max(g[a][b], z);
    }

    cin >> start >> ed;
    dj();
    printf("%.8lf", 100 / d[ed]);
}