POJ 1860 Currency Exchange Bellman-Ford

题意：能否通过套汇盈利。
题解：

#include <iostream>
using namespace std;

double dist[210], v;
int n, m, s;

struct item
{
	int a, b;
	double r,c;
} node[210];

bool Bellman_ford ()
{
	int i, j;
	memset(dist,0,sizeof(dist));
	dist[s] = v;
	for ( i = 1; i < n; i++ )
	{
		for ( j = 1; j <= m * 2; j++ )
			if ( dist[node[j].b] < ( dist[node[j].a] - node[j].c ) * node[j].r )
				dist[node[j].b] = ( dist[node[j].a] - node[j].c ) * node[j].r;
	}
	for ( j = 1; j <= m * 2; j++ )
		if ( dist[node[j].b] < ( dist[node[j].a] - node[j].c ) * node[j].r )
			return true;
	return false;
}

int main()
{
	int a, b;
	double r1, r2, c1, c2;
	cin >> n >> m >> s >> v;
	for ( int i = 1; i <= m * 2; i++ )
	{
		cin >> a >> b >> r1 >> c1 >> r2 >> c2;
		node[i].a = a;
		node[i].b = b;
		node[i].r = r1;
		node[i].c = c1;
		i++;
		node[i].a = b;
		node[i].b = a;
		node[i].r = r2;
		node[i].c = c2;
	}
	
	if ( Bellman_ford () )
		cout << "YES" << endl;
	else
		cout << "NO" << endl;
	return 0;
}