思路：二分+判负环。每次二分一个值mid。推断是否存在小于mid的环，那么就是(w1 + w2 + w3…) / n < mid == w1 – mid + w2 – mid + w3 – mid …. < 0，所以每次二分的时候。把边权值减掉mid。之后bellmanford判负环就可以

代码：

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>#include <queue>using namespace std;typedef double Type;const int MAXNODE = 55;struct Edge {	int u, v;	Type dist;	Edge() {}	Edge(int u, int v, Type dist) {		this->u = u;		this->v = v;		this->dist = dist;	}};struct BellmanFrod {	int n, m;	vector<Edge> edges;	vector<int> g[MAXNODE];	bool inq[MAXNODE];	Type d[MAXNODE];	int p[MAXNODE];	int cnt[MAXNODE];	void init(int n) {		this->n = n;		for (int i = 0; i < n; i++) g[i].clear();		edges.clear();	}	void add_Edge(int u, int v, Type dist) {		edges.push_back(Edge(u, v, dist));		m = edges.size();		g[u].push_back(m - 1);	}	bool negativeCycle() {		queue<int> Q;		memset(inq, 0, sizeof(inq));		memset(cnt, 0, sizeof(cnt));		for (int i = 0; i < n; i++) {			d[i] = 0; inq[i] = 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.v] > d[u] + e.dist) {					d[e.v] = d[u] + e.dist;					p[e.v] = g[u][i];					if (!inq[e.v]) {						Q.push(e.v);						inq[e.v] = true;						if (++cnt[e.v] > n) return true;					}				}			}		}		return false;	}} gao;int T, n, m;bool judge(double mid) {	for (int i = 0; i < gao.m; i++)		gao.edges[i].dist -= mid;	bool tmp = gao.negativeCycle();	for (int i = 0; i < gao.m; i++)		gao.edges[i].dist += mid;	return tmp;}int main() {	scanf("%d", &T);	int cas = 0;	while (T--) {		scanf("%d%d", &n, &m);		gao.init(n);		int u, v;		double dist;		double Max = 0;		while (m--) {			scanf("%d%d%lf", &u, &v, &dist);			u--; v--;			Max = max(Max, dist);			gao.add_Edge(u, v, dist);		}		printf("Case #%d: ", ++cas);		if (!judge(Max + 1)) printf("No cycle found.\n");		else {			double l = 0, r = Max;			for (int i = 0; i < 100; i++) {				double mid = (l + r) / 2;				if (!judge(mid)) l = mid;				else r = mid;			}			printf("%.2lf\n", l);		}	}	return 0;}