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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<string>
 4 #include<cstring>
 5 #include<queue>
 6 #include<map>
 7 #include<vector>
 8 using namespace std;
 9
10 int head[35],nxt[1000],point[1000],size;
11 double val[1000],dist[35];
12 bool vis[35];
13 int n,p,t[35];
14
15 void add(int a,int b,double v){
16     int i;
17     for(i=head[a];~i;i=nxt[i]){
18         if(point[i]==b){
19             if(val[i]<v)val[i]=v;
20             return;
21         }
22     }
23     point[size]=b;
24     val[size]=v;
25     nxt[size]=head[a];
26     head[a]=size++;
27 }
28
29 void spfa(){
30     int i;
31     memset(dist,0,sizeof(dist));
32     memset(vis,0,sizeof(vis));
33     memset(t,0,sizeof(t));
34     dist[1]=100;
35     vis[1]=1;
36     queue<int>q;
37     q.push(1);
38     bool f=1;
39     while(!q.empty()&&f){
40         int u=q.front();
41         q.pop();
42         vis[u]=0;
43         for(i=head[u];~i;i=nxt[i]){
44             int j=point[i];
45             double v=dist[u]*val[i];
46             if(dist[j]<v){
47                 dist[j]=v;
48                 if(!vis[j]){
49                     q.push(j);
50                     vis[j]=1;
51                     t[j]++;
52                     if(t[j]>n)f=0;
53                 }
54             }
55         }
56     }
57     if(f)printf("Case %d: No\n",++p);
58     else printf("Case %d: Yes\n",++p);
59 }
60
61 int main(){
62     p=0;
63     while(scanf("%d",&n)!=EOF&&n!=0){
64         map<string,int>M;
65         int i;
66         string tmp;
67         memset(head,-1,sizeof(head));
68         size=0;
69         for(i=1;i<=n;i++){
70             cin>>tmp;
71             M[tmp]=i;
72         }
73         int m;
74         scanf("%d",&m);
75         for(i=1;i<=m;i++){
76             string t1,t2;
77             double a;
78             cin>>t1>>a>>t2;
79             add(M[t1],M[t2],a);
80         }
81         spfa();
82     }
83     return 0;
84 }