学步园

题意：套汇问题。问能否套汇盈利。
题解：

#include <cstring>
#include <iostream>
using namespace std;

char cur[31][50];
char str1[50], str2[50];
double arb[31][31];
int n,m;

bool floyd ()
{
	int i,j,k;
	for ( k = 0; k < n; k++ )
	{
		for ( i = 0; i < n; i++ )
			for ( j = 0; j < n; j++ )
				if ( arb[i][j] < arb[i][k] * arb[k][j] )
					arb[i][j] = arb[i][k] * arb[k][j];
	}
	for ( i = 0; i < n; i++ )
		if ( arb[i][i] > 1 )
			return true;
	return false;
}

int main()
{
	double num;
	int i, j, x, y, Case = 1;

	while ( cin >> n && n != 0 )
	{
		memset(arb,0,sizeof(arb));
		for ( i = 0; i < n; i++ )
			cin >> cur[i];
		cin >> m;
		for ( i = 0; i < m; i++ )
		{
			cin >> str1 >> num >> str2;
			for ( j = 0; j < n; j++ )
			{
				if ( strcmp(str1,cur[j]) == 0 )
					x = j;
				if ( strcmp(str2,cur[j]) == 0 )
					y = j;
			}
			arb[x][y] = num;
		}
		
		if ( floyd () )
			cout << "Case " << Case++ << ": Yes" << endl;
		else
			cout << "Case " << Case++ << ": No" << endl;
	}
	return 0;
}