还是自己尝试了一下bellmanford，变形之处在于要拿每一个点看是否构成正权回路，即每一个点要用一次bellman-ford()。

#include<iostream>
#include<fstream>
#include<map>

using namespace std;

int n, m;
map<string, int> STL;

char str[50], str1[50], str2[50];
double dist[40], rate;

struct EDGE{
	int ci;
	int cj;
	double cij;
}edges[1000];

bool bellman_ford(int sr)
{
	for (int i = 0; i <= n; i++)
		dist[i] = 0.0;
	dist[sr] = 1.0;
	for (int i = 0; i <n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			if (dist[edges[j].cj] < dist[edges[j].ci] * edges[j].cij)
			{
				dist[edges[j].cj] = dist[edges[j].ci] * edges[j].cij;
			}
		}
	}
	if (dist[sr]>1.0)
		return true;
	else
		return false;
}

int main()
{
	//ifstream in("text.txt");
	int cases = 0;
	while (cin >> n && n != 0)
	{
		cases++;
		for (int i = 1; i <= n; i++)
		{
			cin >> str;
			STL[str] = i;
		}
		cin >> m;
		for (int i = 1; i <= m; i++)
		{
			cin >> str1 >> rate >> str2;
			edges[i].ci = STL[str1];
			edges[i].cj = STL[str2];
			edges[i].cij = rate;
		}
		int sr;
		for (sr= 1; sr <= n; sr++)
		{
			if (bellman_ford(sr))
				break;
		}
		if (sr<=n)
			cout << "Case " << cases << ": Yes" << endl;
		else
			cout << "Case " << cases << ": No" << endl;
	}
	//system("pause");
	return 0;
}