最短路(Floyd_Warshall) POJ 2240 Arbitrage

 

题目传送门

 1 /*  2  最短路：Floyd模板题  3  只要把+改为*就ok了，热闹后判断d[i][i]是否大于1  4  文件输入的ONLINE_JUDGE少写了个_，WA了N遍:)  5 */  6 #include <cstdio>  7 #include <iostream>  8 #include <cstring>  9 #include <algorithm> 10 #include <string> 11 #include <map> 12 #include <cmath> 13 #include <vector> 14 #include <set> 15 #include <queue> 16 using namespace std; 17 18 const int MAXN = 1e6 + 10; 19 const int INF = 0x3f3f3f3f; 20 double d[33][33]; 21 22 void Floyd_Warshall(int n) 23 { 24 for (int k=1; k<=n; ++k) 25  { 26 for (int i=1; i<=n; ++i) 27  { 28 for (int j=1; j<=n; ++j) 29  { 30 if (d[i][j] < d[i][k] * d[k][j]) 31  { 32 d[i][j] = d[i][k] * d[k][j]; 33  } 34  } 35  } 36  } 37 38 for (int i=1; i<=n; ++i) 39  { 40 if (d[i][i] > 1) 41  { 42 puts ("Yes"); return ; 43  } 44  } 45 46 puts ("No"); return ; 47 } 48 49 int main(void) //POJ 2240 Arbitrage 50 { 51  #ifndef ONLINE_JUDGE 52 freopen ("F.in", "r", stdin); 53 #endif 54 55 int n, num; int cas = 0; 56 while (cin >> n && n) 57  { 58 for (int i=1; i<=n; ++i) 59  { 60 for (int j=1; j<=n; ++j) 61  { 62 if (i == j) d[i][j] = 1; 63 else d[i][j] = 0; 64  } 65  } 66 67 map<string, int> m; 68 string s, s1, s2; 69 for (int i=1; i<=n; ++i) 70  { 71 cin >> s; 72 m[s] = i; 73  } 74 75 cin >> num; 76 for (int i=1; i<=num; ++i) 77  { 78 double w; 79 cin >> s1 >> w >> s2; 80 d[m[s1]][m[s2]] = w; 81  } 82 83 cout << "Case " << ++cas << ": "; 84  Floyd_Warshall (n); 85  } 86 87 return 0; 88 }