会建图，然后使用标准的Bellman Ford算法，判断负环就解决了。

不过本题实际应用不是计算负环，而是计算最大值，也就是求出源点到所有点的最大收益值之后，然后判断是否可以进一步增加收益，如果可以那么证明有环可以不断反复走这个环，不断增加收益，实际就是判负环的应用了。

#include <stdio.h>
#include <vector>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <limits.h>
#include <stack>
#include <queue>
#include <set>
#include <map><span style="white-space:pre">	</span>
#include <float.h>
using namespace std;

const int MAX_N = 101;
int N, M, Src;
double V, dis[MAX_N];
struct Node
{
	int s, d;//source and destination vertex
	double w, c;//exchange rate and commission
};

Node gra[MAX_N<<1];

bool BellmanFord(int src, int n, double val)//n: number of edges
{
	fill(dis, dis+N+1, 0.0);
	dis[src] = val;
	for (int i = 1; i < N; i++)//标准Bellman Ford Algorithm
	{
		for (int j = 0; j < n; j++)
		{
			int u = gra[j].s, v = gra[j].d;
			if (dis[u] != 0.0)
			{
				double t = (dis[u]-gra[j].c) * gra[j].w;
				if (t > dis[v]) dis[v] = t;
			}
		}
	}

	for (int j = 0; j < n; j++)//判断负环
	{
		int u = gra[j].s, v = gra[j].d;
		if (dis[u] != 0.0)
		{
			double t = (dis[u] - gra[j].c) * gra[j].w;
			if (t > dis[v]) return true;
		}
	}
	return false;
}
int main()
{
	while (scanf("%d %d %d %lf", &N, &M, &Src, &V) != EOF)
	{
		int u, v, id = 0;
		double a, b, c, d;
		for (int i = 0; i < M; i++)
		{
			scanf("%d %d %lf %lf %lf %lf", &u, &v, &a, &b, &c, &d);
			gra[id].s = u, gra[id].d = v;
			gra[id].w = a, gra[id].c = b;
			id++;
			gra[id].s = v, gra[id].d = u;
			gra[id].w = c, gra[id].c = d;
			id++;
		}
		if (BellmanFord(Src, id, V)) puts("YES");
		else puts("NO");
	}
	return 0;
}

POJ 1860 Currency Exchange BellmanFord题解