POJ1860-Currency Exchange-判正环

两种货币的交换可以当成两条边，建图后跑Bellman_ford算法就好了。

Bellman_ford算法可以用来处理负边权，所以可以判断是否存在负环。反过来就可以判断是否存在正环。

 /*--------------------------------------------------------------------------------------*/
 //        Helica's header
 //        Second Edition
 //        2015.11.7
 //
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <ctype.h>
 #include <cstdlib>
 #include <cstdio>
 #include <vector>
 #include <string>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <set>
 #include <map>

 //debug function for a N*M array
 #define debug_map(N,M,G) printf("\n");for(int i=0;i<(N);i++)\
 {for(int j=;j<(M);j++){\
 printf("%d",G[i][j]);}printf("\n");}
 //debug function for int,float,double,etc.
 #define debug_var(X) cout<<#X"="<<X<<endl;
 /*--------------------------------------------------------------------------------------*/
 using namespace std;

 const int maxn = +;
 int N,M,T,S;
 int tol;
 double V;

 struct Edge
 {
     int u,v;
     double c,r;
 }E[*maxn];

 double dist[maxn];

 bool bellman_ford(int start,int n)
 {
     memset(dist,,sizeof dist);
     dist[start] = V;

     for(int i=;i<n;i++)
     {
         bool flag = false;
         for(int j=;j<tol;j++)if(dist[E[j].v] < (dist[E[j].u]-E[j].c)*E[j].r )
         {
             dist[E[j].v] = (dist[E[j].u]-E[j].c)*E[j].r;
            flag = true;
         }
         if(!flag) return false;
     }

     for(int j=;j<tol;j++)
     {
         if(dist[E[j].v] < (dist[E[j].u]-E[j].c)*E[j].r)
             return true;
     }
     return false;
 }

 int main()
 {
     while(~scanf("%d%d%d%lf",&N,&M,&S,&V))
     {
         int a,b;
         double r1,r2,c1,c2;
         tol = ;
         for(int i=;i<M;i++)
         {
             scanf("%d%d%lf%lf%lf%lf",&a,&b,&r1,&c1,&r2,&c2);
             E[tol].u = a;E[tol].v = b;
             E[tol].c = c1;E[tol++].r = r1;
             E[tol].u = b;E[tol].v = a;
             E[tol].c = c2;E[tol++].r = r2;
         }
         if(bellman_ford(S,N))
             printf("YES\n");
         else
             printf("NO\n");

     }
 }