hdu 4463 有一条边必须加上 （2012杭州区域赛K题）

耐克店 和 苹果店必须相连

Sample Input
4
2 3
0 0
1 0
0 -1
1 -1
0

Sample Output
3.41

 # include <iostream>
 # include <cstdio>
 # include <cstring>
 # include <algorithm>
 # include <cmath>
 # define LL long long
 using namespace std ;

 const int INF=0x3f3f3f3f;
 const int MAXN=;
 bool vis[MAXN];
 double lowc[MAXN];
 int n ;
 double cost[MAXN][MAXN] ;

 struct poin
 {
     int x ;
     int y ;
 }p[MAXN];

 double Prim()//点是0~n-1
 {
     double ans=;
     memset(vis,false,sizeof(vis));
     vis[]=true;
     for(int i=;i<n;i++)lowc[i]=cost[][i];
     for(int i=;i<n;i++)
     {
         double minc=INF;
         int p=-;
         for(int j=;j<n;j++)
             if(!vis[j]&&minc>lowc[j])
             {
                 minc=lowc[j];
                 p=j;
             }
             if(minc==INF)return -;//原图不连通
             ans+=minc;
             vis[p]=true;
             for(int j=;j<n;j++)
                 if(!vis[j]&&lowc[j]>cost[p][j])
                     lowc[j]=cost[p][j];
     }
     return ans;
 }

 int main()
 {

    // freopen("in.txt","r",stdin) ;
     while(scanf("%d" , &n) != EOF)
     {
         if (n == )
             break ;
         int i , j ;
         int pp , qq ;
         scanf("%d %d" , &pp , &qq) ;
         for (i =  ; i < n ; i++)
             scanf("%d %d" , &p[i].x , &p[i].y) ;
         for (i =  ; i < n ; i++)
             for (j = i+ ; j < n ; j++)
             {
                 double t = sqrt((double)(p[i].x - p[j].x) * (p[i].x - p[j].x) + (p[i].y - p[j].y) * (p[i].y - p[j].y)) ;
                 cost[i][j] = t ;
                 cost[j][i] = t ;
             }
         double k = cost[pp-][qq-] ;
         cost[pp-][qq-] =  ;
         cost[qq-][pp-] =  ;
         k += Prim() ;
         printf("%.2lf\n" , k) ;

     }
     return  ;
 }