poj2240 最短路判环

题意：与poj1680一样，有不同的换钱渠道，可以完成特定两种货币的交换，并且有汇率，只不过此题是单向边，然后问是否能使财富增加

与poj1680一样，建图之后直接spfa判增值的环即可

 #include<iostream>
 #include<cstdio>
 #include<string>
 #include<cstring>
 #include<queue>
 #include<map>
 #include<vector>
 using namespace std;

 int head[],nxt[],point[],size;
 double val[],dist[];
 bool vis[];
 int n,p,t[];

 void add(int a,int b,double v){
     int i;
     for(i=head[a];~i;i=nxt[i]){
         if(point[i]==b){
             if(val[i]<v)val[i]=v;
             return;
         }
     }
     point[size]=b;
     val[size]=v;
     nxt[size]=head[a];
     head[a]=size++;
 }

 void spfa(){
     int i;
     memset(dist,,sizeof(dist));
     memset(vis,,sizeof(vis));
     memset(t,,sizeof(t));
     dist[]=;
     vis[]=;
     queue<int>q;
     q.push();
     bool f=;
     while(!q.empty()&&f){
         int u=q.front();
         q.pop();
         vis[u]=;
         for(i=head[u];~i;i=nxt[i]){
             int j=point[i];
             double v=dist[u]*val[i];
             if(dist[j]<v){
                 dist[j]=v;
                 if(!vis[j]){
                     q.push(j);
                     vis[j]=;
                     t[j]++;
                     if(t[j]>n)f=;
                 }
             }
         }
     }
     if(f)printf("Case %d: No\n",++p);
     else printf("Case %d: Yes\n",++p);
 }

 int main(){
     p=;
     while(scanf("%d",&n)!=EOF&&n!=){
         map<string,int>M;
         int i;
         string tmp;
         memset(head,-,sizeof(head));
         size=;
         for(i=;i<=n;i++){
             cin>>tmp;
             M[tmp]=i;
         }
         int m;
         scanf("%d",&m);
         for(i=;i<=m;i++){
             string t1,t2;
             double a;
             cin>>t1>>a>>t2;
             add(M[t1],M[t2],a);
         }
         spfa();
     }
     return ;
 }