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。

 1 /*--------------------------------------------------------------------------------------*/
 2 //        Helica's header 
 3 //        Second Edition
 4 //        2015.11.7
 5 //
 6 #include <algorithm>
 7 #include <iostream>
 8 #include <cstring>
 9 #include <ctype.h>
10 #include <cstdlib>
11 #include <cstdio>
12 #include <vector>
13 #include <string>
14 #include <queue>
15 #include <stack>
16 #include <cmath>
17 #include <set>
18 #include <map>
19 
20 //debug function for a N*M array 
21 #define debug_map(N,M,G) printf("\n");for(int i=0;i<(N);i++)\
22 {for(int j=0;j<(M);j++){\
23 printf("%d",G[i][j]);}printf("\n");}                
24 //debug function for int,float,double,etc.
25 #define debug_var(X) cout<<#X"="<<X<<endl;
26 /*--------------------------------------------------------------------------------------*/
27 using namespace std;
28 
29 const int maxn = 100+10;
30 int N,M,T,S;
31 int tol;
32 double V;
33 
34 struct Edge
35 {
36     int u,v;
37     double c,r;
38 }E[10*maxn];
39 
40 double dist[maxn];
41 
42 bool bellman_ford(int start,int n)
43 {
44     memset(dist,0,sizeof dist);
45     dist[start] = V;
46 
47     for(int i=1;i<n;i++)
48     {
49         bool flag = false;
50         for(int j=0;j<tol;j++)if(dist[E[j].v] < (dist[E[j].u]-E[j].c)*E[j].r )
51         {
52             dist[E[j].v] = (dist[E[j].u]-E[j].c)*E[j].r;
53            flag = true;    
54         }
55         if(!flag) return false;
56     }
57 
58     for(int j=0;j<tol;j++)
59     {
60         if(dist[E[j].v] < (dist[E[j].u]-E[j].c)*E[j].r)
61             return true;
62     }
63     return false;
64 }
65 
66 int main()
67 {
68     while(~scanf("%d%d%d%lf",&N,&M,&S,&V))
69     {
70         int a,b;
71         double r1,r2,c1,c2;
72         tol = 0;
73         for(int i=0;i<M;i++)
74         {
75             scanf("%d%d%lf%lf%lf%lf",&a,&b,&r1,&c1,&r2,&c2);
76             E[tol].u = a;E[tol].v = b;
77             E[tol].c = c1;E[tol++].r = r1;
78             E[tol].u = b;E[tol].v = a;
79             E[tol].c = c2;E[tol++].r = r2;
80         }
81         if(bellman_ford(S,N))
82             printf("YES\n");
83         else
84             printf("NO\n");
85 
86     }    
87 }