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。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <queue> 6 #include <string> 7 #include <map> 8 #include <cmath> 9 #include <iomanip> 10 using namespace std; 11 12 typedef long long LL; 13 #define inf (1LL << 25) 14 #define rep(i,j,k) for(int i = (j); i <= (k); i++) 15 #define rep__(i,j,k) for(int i = (j); i < (k); i++) 16 #define per(i,j,k) for(int i = (j); i >= (k); i--) 17 #define per__(i,j,k) for(int i = (j); i > (k); i--) 18 19 const int N = 40; 20 map<string,int > si; //编号，方便建图 21 int head[N]; 22 bool vis[N]; 23 int tot[N]; 24 double value[N]; 25 int cnt; 26 int n; 27 28 struct Edge{ 29 int to; 30 double w; 31 int next; 32 }e[1010]; 33 34 void add(int u,int v,double w){ 35 e[cnt].to = v; 36 e[cnt].w = w; 37 e[cnt].next = head[u]; 38 head[u] = cnt++; 39 } 40 41 bool SPFA(){ 42 43 rep(i,1,n) value[i] = 0; 44 value[1] = 1000; //随意选个点就行 45 rep(i,1,n) vis[i] = false; 46 rep(i,1,n) tot[i] = 0; 47 vis[1] = true; 48 queue<int> que; 49 que.push(1); 50 51 while(!que.empty()){ 52 int u = que.front(); 53 que.pop(); 54 vis[u] = false; 55 56 for(int o = head[u]; ~o; o = e[o].next){ 57 int v = e[o].to; 58 double w = e[o].w; 59 60 if(value[v] < value[u] * w){ 61 value[v] = value[u] * w; 62 if(!vis[v]){ 63 vis[v] = true; 64 que.push(v); 65 tot[v]++; 66 if(tot[v] > n - 1) return true; //有正环 67 } 68 } 69 } 70 } 71 72 return false; 73 74 } 75 76 int main(){ 77 78 ios::sync_with_stdio(false); 79 cin.tie(0); 80 81 int tot = 0; 82 while(cin >> n && n){ 83 84 rep(i,1,n) head[i] = -1; 85 cnt = 0; 86 si.clear(); 87 string in; 88 rep(i,1,n){ 89 cin >> in; 90 si[in] = i; 91 } 92 93 int m; 94 string u,v; 95 double w; 96 97 cin >> m; 98 rep(i,1,m){ 99 cin >> u >> w >> v; 100 add(si[u],si[v],w); 101 } 102 cout << "Case " << ++tot << ": "; 103 if(SPFA()) cout << "Yes" << endl; 104 else cout << "No" << endl; 105 } 106 107 108 109 getchar(); getchar(); 110 return 0; 111 }