Arbitrage POJ - 2240

题目链接：https://vjudge.net/problem/POJ-2240

思路：判正环，Bellman-ford和SPFA,floyd都可以，有正环就可以套利。

这里用SPFA，就是个板子题吧，把松弛改成乘法操作就好了。

---

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <queue>
 #include <string>
 #include <map>
 #include <cmath>
 #include <iomanip>
 using namespace std;
 
 typedef long long LL;
 #define inf (1LL << 25)
 #define rep(i,j,k) for(int i = (j); i <= (k); i++)
 #define rep__(i,j,k) for(int i = (j); i < (k); i++)
 #define per(i,j,k) for(int i = (j); i >= (k); i--)
 #define per__(i,j,k) for(int i = (j); i > (k); i--)
 
 const int N = ;
 map<string,int > si; //编号，方便建图
 int head[N];
 bool vis[N];
 int tot[N];
 double value[N];
 int cnt;
 int n;
 
 struct Edge{
     int to;
     double w;
     int next;
 }e[];
 
 void add(int u,int v,double w){
     e[cnt].to = v;
     e[cnt].w = w;
     e[cnt].next = head[u];
     head[u] = cnt++;
 }
 
 bool SPFA(){
 
     rep(i,,n) value[i] = ;
     value[] = ; //随意选个点就行
     rep(i,,n) vis[i] = false;
     rep(i,,n) tot[i] = ;
     vis[] = true;
     queue<int> que;
     que.push();
 
     while(!que.empty()){
         int u = que.front();
         que.pop();
         vis[u] = false;
 
         for(int o = head[u]; ~o; o = e[o].next){
             int v = e[o].to;
             double w = e[o].w;
 
             if(value[v] < value[u] * w){
                 value[v] = value[u] * w;
                 if(!vis[v]){
                     vis[v] = true;
                     que.push(v);
                     tot[v]++;
                     if(tot[v] > n - ) return true; //有正环
                 }
             }
         }
     } 
 
     return false;
 
 }
 
 int main(){
 
     ios::sync_with_stdio(false);
     cin.tie();
 
     int tot = ;
     while(cin >> n && n){
 
         rep(i,,n) head[i] = -;
         cnt = ;
         si.clear();
         string in;
         rep(i,,n){
             cin >> in;
             si[in] = i;
         }
 
         int m;
         string u,v;
         double w;
 
         cin >> m;
         rep(i,,m){
             cin >> u >> w >> v;
             add(si[u],si[v],w);
         }
         cout << "Case " << ++tot << ": ";
         if(SPFA()) cout << "Yes" << endl;
         else cout << "No" << endl;
     }
 
     getchar(); getchar();
     return ;
 }