題目大意：
給你一個由n個點，m條邊組成的有向圖，求圖中最小環的平均長度。
分析：
根據最小環的條件，我們可以通過二分的方法來實現，對於每個當前二分出來的答案，要使得環上的每一條邊的平均值小於當前答案，就等價於

#include<bits/stdc++.h>
using namespace std;
const int maxn = 60;

struct
 
 Edge { int from, to; double dist; };
struct BellmanFord {
    int n, m;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool inque[maxn];
    double dis[maxn];
    int pre[maxn], cnt[maxn];

    void init(int n) {
        this->n = n;
        for(int i=0; i<n; i++) G[i].clear();
        edges.clear();
    }
    void
 
 AddEdge(int from, int to, double dist) {
        edges.push_back((Edge){from, to, dist});
        m = edges.size();
        G[from].push_back(m-1);
    }
    bool negativeCycle() {
        queue<int> q;
        memset(cnt, 0, sizeof(cnt));
        memset(inque, false, sizeof(inque));

        for
 
(int i=0; i<n; i++) dis[i] = 0, q.push(i), inque[i] = true;
        while(!q.empty()) {
            int u = q.front(); q.pop(); inque[u] = false;
            for(int i=0; i<(int)G[u].size(); i++) {
                Edge& e = edges[G[u][i]];

                if(dis[e.to] > dis[u] + e.dist) {
                    dis[e.to] = dis[u] + e.dist;
                    if(!inque[e.to]) {
                        q.push(e.to);
                        pre[e.to] = G[u][i];
                        inque[e.to] = true;
                        if(++cnt[e.to] > n) return true;
                    }
                }
            }
        }
        return false;
    }
}Bel;

bool check(double mid) {
    for(int i=0; i<Bel.m; i++) 
        Bel.edges[i].dist -= mid;
    bool res = Bel.negativeCycle();
    for(int i=0; i<Bel.m; i++) 
        Bel.edges[i].dist += mid;
    return res;
}
double binary_search(double L, double R) {
    while(R - L > 1e-5) {
        double M = L + (R-L)/2;
        if(check(M)) R = M; else L = M;
    }
    return L;
}
int Case, T, n, m, u, v, w;
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    freopen("ans.txt", "w", stdout);
#endif
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);

        Bel.init(n);
        double L = 0, R = 0;
        for(int i=0; i<m; i++) {
            scanf("%d%d%d", &u, &v, &w); 
            u--, v--, R = R < w ? w : R;
            Bel.AddEdge(u, v, w);
        }
        printf("Case #%d: ", ++Case);
        if(!check(R + 1)) { printf("No cycle found.\n"); continue; } 
        double ans = binary_search(L, R);
        printf("%.2f\n", ans);
    }
    return 0;
}