最短路 POJ 2240 Arbitrage

 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 }