最少手续费,跑最短路即可注意精度
#include <bits/stdc++.h>
using namespace std;
const int N = 2010;
const int M = 1e5 + 10;
double g[N][N];
double dist[N];
bool st[N];
int n, m;
int sta, ed;
void dijkstra()
{
dist[sta] = 1;
for (int i = 0; i < n; i++)
{
int t = -1;
for (int j = 1; j <= n; j++)
{
if (!st[j] && (t == -1 || dist[t] < dist[j]))
t = j;
}
st[t] = true;
for (int j = 1; j <= n; j++)
dist[j] = max(dist[j], dist[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 - c) * 1.0 / 100;
g[a][b] = g[b][a] = max(g[a][b], z);
}
cin >> sta >> ed;
dijkstra();
printf("%.8f", 100 / dist[ed]);
return 0;
}