poj2240 floyd

 1 //Accepted    732 KB    782 ms
 2 //floyd应用
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <iostream>
 6 #include <queue>
 7 #include <cmath>
 8 #include <map>
 9 #include <algorithm>
10 using namespace std;
11 /**
12   * This is a documentation comment block
13   * 如果有一天你坚持不下去了，就想想你为什么走到这儿！
14   * @authr songt
15   */
16 const int imax_n = 35;
17 double a[imax_n][imax_n];
18 map<string ,int > mp;
19 int n;
20 double max(double a,double b)
21 {
22     if (a-b<1e-9) return b;
23     return a;
24 }
25 void floyd()
26 {
27     for (int k=1;k<=n;k++)
28     for (int i=1;i<=n;i++)
29     for (int j=1;j<=n;j++)
30     if (a[i][k]>1e-9 && a[k][j]>1e-9)
31     a[i][j]=max(a[i][j],a[i][k]*a[k][j]);
32     /*
33     for (int i=1;i<=n;i++)
34     {
35         for (int j=1;j<=n;j++)
36         {
37             printf("%.3lf ",a[i][j]);
38         }
39         printf("\n");
40     }
41     */
42 }
43 int main()
44 {
45     int t=0;
46     string s;
47     string s1,s2;
48     double rate;
49     while (scanf("%d",&n),n)
50     {
51         mp.clear();
52         for (int i=1;i<=n;i++)
53         {
54             cin>>s;
55             mp[s]=i;
56         }
57         int m;
58         scanf("%d",&m);
59         for (int i=1;i<=n;i++)
60         for (int j=1;j<=n;j++)
61         {
62             if (i==j) a[i][j]=1;
63             else a[i][j]=0;
64         }
65         for (int i=1;i<=m;i++)
66         {
67             cin>>s1>>rate>>s2;
68             //cout<<mp[s1]<<" "<<mp[s2]<<endl;
69             a[mp[s1]][mp[s2]]=rate;
70         }
71         floyd();
72         int flag=0;
73         for (int i=1;i<=n;i++)
74         if (a[i][i]-1>1e-9) flag=1;
75         printf("Case %d: ",++t);
76         if (flag) printf("Yes\n");
77         else printf("No\n");
78     }
79     return 0;
80 }