poj2253

 1 //Accepted    468 KB    16 ms
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <queue>
 6 #include <cmath>
 7 #include <algorithm>
 8 using namespace std;
 9 /**
10   * This is a documentation comment block
11   * 如果有一天你坚持不下去了，就想想你为什么走到这儿！
12   * @authr songt
13   */
14 const int imax_n = 205;
15 const int inf = 100000000;
16 int x[imax_n];
17 int y[imax_n];
18 double a[imax_n][imax_n];
19 bool vis[imax_n];
20 double dis[imax_n];
21 int n;
22 double max(double a,double b)
23 {
24     if (a-b<1e-9) return b;
25     return a;
26 }
27 double min(double a,double b)
28 {
29     if (a-b>1e-9) return b;
30     return a;
31 }
32 void dij(int src)
33 {
34     memset(vis,false,sizeof(vis));
35     vis[src]=true;
36     for (int i=1;i<=n;i++)
37     dis[i]=a[src][i];
38     for (int i=1;i<=n;i++)
39     {
40         double temp=inf;
41         int k=-1;
42         for (int j=1;j<=n;j++)
43         {
44             if (vis[j]) continue;
45             if (temp-dis[j]>1e-9)
46             {
47                 temp=dis[j];
48                 k=j;
49             }
50         }
51         if (k==-1) return ;
52         vis[k]=true;
53         for (int j=1;j<=n;j++)
54         if (!vis[j])
55         {
56             dis[j]=min(dis[j],max(dis[k],a[k][j]));
57         }
58     }
59 }
60 int main()
61 {
62     int t=0;
63     while (scanf("%d",&n),n)
64     {
65         for (int i=1;i<=n;i++)
66         {
67             scanf("%d%d",&x[i],&y[i]);
68         }
69         for (int i=1;i<=n;i++)
70         {
71             for (int j=1;j<=n;j++)
72             a[i][j]=sqrt((double )(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
73         }
74         dij(1);
75         printf("Scenario #%d\nFrog Distance = %.3lf\n\n",++t,dis[2]+0.000005);
76     }
77     return 0;
78 }