POJ 1860 Currency Exchange Bellma求有无环 .

题目地址：http://poj.org/problem?id=1860

正好是Bellman的逆运算

求的是最长路径，且如果有一条能一直走的正环(代表资产一直增加)就输出YES

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int INF=(1<<30);
struct Edge{int from,to; double r,c;Edge(int f,int t,double r,double c):from(f),to(t),r(r),c(c){}
};
vector<Edge> edges;
double d[100+5];
bool Bellman_ford(int s,double val,int n)
{fill(d+1,d+n+1,0);d[s]=val;for(int k=1;k<n;k++) //从u~v点经过k条边for(int i=0;i<edges.size();i++){int u=edges[i].from;int v=edges[i].to;double r=edges[i].r, c=edges[i].c;if(d[v]<(d[u]-c)*r) d[v]=(d[u]-c)*r;}for(int i=0;i<edges.size();i++){int u=edges[i].from;int v=edges[i].to;double r=edges[i].r, c=edges[i].c;if(d[v]<(d[u]-c)*r) return true;} return false;
}
int main()
{int n,m,s; double val;cin>>n>>m>>s>>val;int u,v; double r,c;for(int i=0;i<m;i++){cin>>u>>v>>r>>c;edges.push_back(Edge(u,v,r,c));cin>>r>>c;edges.push_back(Edge(v,u,r,c));}cout<<(Bellman_ford(s,val,n)?"YES":"NO")<<endl;return 0;
}