POJ 2253

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<math.h>
 4 #include<iomanip>
 5 #define MAXN 250
 6 #define inf 1000000000
 7 typedef double elem_t;
 8 
 9 using namespace std;
10 
11 double _m[MAXN][MAXN];
12 struct point
13 {
14     double a;
15     double b;
16 };
17 elem_t prim(int n,elem_t mat[][MAXN],int* pre);
18 point p[250];
19 int pre[250];
20 int main()
21 {
22     //freopen("acm.acm","r",stdin);
23     int num;
24     int max = 0;
25     int i;
26     int time = 0;
27     int j;
28     int value;
29     while(cin>>num)
30     {
31         if(!num)
32             break;
33         cin>>p[0].a;
34         cin>>p[0].b;
35         cin>>p[1].a;
36         cin>>p[1].b;
37         memset(_m,0,sizeof(_m));
38         for(i = 2; i < num; ++ i)
39         {
40             cin>>p[i].a;
41             cin>>p[i].b;
42         }
43         for(i = 0; i < num; ++ i)
44         {
45             for(j = i; j < num; ++ j)
46             {
47                 _m[i][j] = (p[i].a - p[j].a)*(p[i].a - p[j].a) + (p[i].b - p[j].b)*(p[i].b - p[j].b);
48                 _m[j][i] = _m[i][j];
49             }
50         }
51         prim(num,_m,pre);
52         i = 1;
53         while(1)
54         {
55             if(_m[i][pre[i]] > max)
56             {
57                 max = _m[i][pre[i]];
58             }
59             if(pre[i] == 0)
60                 break;
61             i = pre[i];
62         }
63         cout<<"Scenario #"<<++ time<<endl
64             <<"Frog Distance = "
65             <<setiosflags(ios::fixed)<<setprecision(3)<<sqrt(long double (max))<<endl
66             <<endl;
67         max = 0;
68     }
69 
70 }
71 
72 
73 
74 elem_t prim(int n,elem_t mat[][MAXN],int* pre){
75     elem_t min[MAXN],ret=0;
76     int v[MAXN],i,j,k;
77     for (i=0;i<n;i++)
78         min[i]=inf,v[i]=0,pre[i]=-1;
79     for (min[j=0]=0;j<n;j++){
80         for (k=-1,i=0;i<n;i++)
81             if (!v[i]&&(k==-1||min[i]<min[k]))
82                 k=i;
83         for (v[k]=1,ret+=min[k],i=0;i<n;i++)
84             if (!v[i]&&mat[k][i]<min[i])
85                 min[i]=mat[pre[i]=k][i];
86     }
87     return ret;
88 }