Description

I IU P C2 0 06
Problem G: Going in Cycle!!
Input: standard input Output: standard output
You are given a weighted directed graph with n vertices and m edges. Each cycle in the graph has a weight, which equals to sum of its edges. There are so many cycles in the graph with different weights. In this problem we want to find a cycle with the minimum mean.
Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with two numbers n andm. m lines follow, each has three positive numbera, b, c which means there is an edge from vertex a tob with weight of c.
Output
For each test case output one line containing “Case #x: ” followed by a number that is the lowest mean cycle in graph with 2 digits after decimal place, if there is a cycle. Otherwise print “No cycle found.”.
Constraints
-           n ≤ 50 -           a, b ≤ n -           c ≤ 10000000
Sample Input	Output for Sample Input
2 2 1 1 2 1 2 2 1 2 2 2 1 3	Case #1: No cycle found. Case #2: 2.50

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：先使用二分求解mid。假设存在一个包含k条边的回路，回路上各个边的权值为w1,w2....wk,那么平均值小于mid

意味着：w1+w2..+wk<K*mid -> (w1-mid)+(w2-mid)+...(wk-mid) < 0，就是判断新图中是否含有负权回路了，两种初始化结果都对

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 60;
const int inf = 0x3f3f3f3f;

struct Edge {
	int from, to;
	double dist;
};

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 AddEdge(int from, int to, int dist) {
		edges.push_back((Edge){from, to, dist});
		m = edges.size();
		G[from].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[0] = 1;
			Q.push(i);
		}
		*/
		for (int i = 0; i < n; i++) {
			d[i] = 0;
			inq[i] = 1;
			Q.push(i);
			cnt[i] = 1;
		}

		while (!Q.empty()) {
			int u = Q.front();
			Q.pop();
			inq[u] = 0;
			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] = 1;
						if (++cnt[e.to] > n)
							return 1;
					}
				}
			}
		}
		return 0;
	}
};
BellmanFord solver;

bool test(double x) {
	for (int i = 0; i < solver.m; i++)
		solver.edges[i].dist -= x;
	bool ret = solver.negativeCycle();
	for (int i = 0; i < solver.m; i++)
		solver.edges[i].dist += x;
	return ret;
}


int main() {
	int t;
	scanf("%d", &t);
	for (int cas = 1; cas <= t; cas++) {
		int n, m;
		scanf("%d%d", &n, &m);
		solver.init(n);
		int ub = 0;
		while (m--) {
			int u, v, w;
			scanf("%d%d%d", &u, &v, &w);
			u--, v--;
			ub = max(ub, w);
			solver.AddEdge(u, v, w);
		}
		printf("Case #%d: ", cas);
		if (!test(ub+1))
			printf("No cycle found.\n");
		else {
			double L = 0, R = ub;
			while (R-L > 1e-3) {
				double M = L + (R-L)/2;
				if (test(M))
					R = M;
				else L = M;
			}
			printf("%.2lf\n", L);
		}
	}
	return 0;
}