题解里大家都是从终点推起点的做法，但是我一开始做的时候觉得正着推二分可行就直接莽了没多想2333
效率还是挺低的，但是能ac，注意二分的精度要拉到1e-10，不然部分样例会挂掉

#include <bits/stdc++.h>
using namespace std;
const int N = 2005, M = 2e5 + 5;
int head[N], ne[M], edge[M], wg[M], idx, n, m;
double dist[N];
bool st[N];
struct HeapNode
{
    int id;
    double val, cost;
    bool operator<(const HeapNode& t) const
    {
        return cost > t.cost;
    }
};
void add(int u, int v, int w)
{
    edge[idx] = v;
    wg[idx] = w;
    ne[idx] = head[u];
    head[u] = idx++;
}
void dijkstra(int x, double val)
{
    memset(dist, 0x7f, sizeof(dist));
    memset(st, false, sizeof(st));
    dist[x] = 0;
    priority_queue<HeapNode> q;
    q.push({ x, val, 0});
    while (!q.empty()) {
        HeapNode node = q.top();
        q.pop();
        if (st[node.id]) continue;
        for (int j = head[node.id]; j != -1; j = ne[j])
        {
            double num = node.val * wg[j] * 0.01;
            if (!st[edge[j]] && num + node.cost < dist[edge[j]]) {
                dist[edge[j]] = num + node.cost;
                q.push({ edge[j], node.val - num, dist[edge[j]] });
            }
        }
        st[node.id] = true;
    }
}
int main()
{
    memset(head, -1, sizeof(head));
    cin >> n >> m;
    for (int i = 0; i < m; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        add(u, v, w);
        add(v, u, w);
    }
    int a, b;
    cin >> a >> b;
    double l = 100, r = 1e9, ans;
    while (r - l > 1e-10) {
        double mid = (l + r) / 2;
        dijkstra(a, mid);
        if (mid - dist[b] >= 100) {
            ans = mid;
            r = mid;
        }
        else l = mid;
    }
    printf("%.8f\n", ans);
    return 0;
}