题意：给定一个n个点m条边的加权有向图，求平均权值最小的回路。

思路：使用二分法求解，对于每一个枚举值mid，判断每条边权值减去mid后有无负圈即可。

#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<cstdlib>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<queue>  
#include<stack> 
#include<string>
#include<map> 
#include<set>
#define eps 1e-4 
#define LL long long  
using namespace std;  

const int maxn = 100 + 5;
const int INF = 0x3f3f3f3f;

//Bellman-Ford算法
struct Edge {
	int from, to;
	double dist;
	Edge(int u = 0, int v = 0, double d = 0) : from(u), to(v), dist(d) {
	}
};
struct BellmanFord{
	int n, m;
	vector<Edge> edges;
	vector<int> G[maxn];
	bool inq[maxn];
	double d[maxn];
	int p[maxn];
	int cnt[maxn];
	
	void init(int n) {
		this->n = n;
		for(int i = 0; i < n; i++) G[i].clear();
		edges.clear();
	}
	
	void AddEdges(int from, int to, int dist) {
		edges.push_back(Edge(from, to, dist));
		m = edges.size();
		G[from].push_back(m-1);
	}
	
	bool negetiveCycle() {
		queue<int> Q;
		memset(inq, 0, sizeof(inq));
		memset(cnt, 0, sizeof(cnt));
		for(int i = 0; i < n; i++) {
			d[i] = 0; inq[0] = true; Q.push(i);
		}
		while(!Q.empty()) {
			int u = Q.front(); Q.pop();
			inq[u] = false;
			for(int i = 0; i < G[u].size(); i++) {
				Edge& e = edges[G[u][i]];
				if(d[e.to]>d[u]+e.dist) {
					d[e.to] = d[u] + e.dist;
					p[e.to] = G[u][i];
					if(!inq[e.to]) {
						Q.push(e.to);
						inq[e.to] = true;
						if(++cnt[e.to] > n) return true;
					}
				}
			}
		}
		return false;
	}
	
} solver; 

int n, m;
bool test(double w) {
	for(int i = 0; i < m; i++) solver.edges[i].dist -= w;
	int t = solver.negetiveCycle();
	for(int i = 0; i < m; i++) solver.edges[i].dist += w;
	return t;
}

int kase;
int main() {
	freopen("input.txt", "r", stdin);
	int t; cin >> t;
	while(t--) {
		cin >> n >> m;
		solver.init(n);
		for(int i = 0; i < m; i++) {
			int u, v, d;
			cin >> u >> v >> d;
			u--; v--;
			solver.AddEdges(u, v, d);
		}
		double L = -10000001, R = 10000001;
		if(!test(R)) printf("Case #%d: No cycle found.\n", ++kase);
		else {
			while(R-L>eps) {
				double M = (R+L)/2;
				if(test(M)) R = M;
				else L = M;
			}
			printf("Case #%d: %.2lf\n", ++kase, R);
		}
	}
	return 0;
}