hdu 4463 Outlets 解题报告

题意：给定你二维平面内的n个点，有两个点之间必须建路，问你所有点连接起来的最短路径是多少

解题思路：裸的最小生成树，不过在实现最小生成树之前，应该初始它的并查集和临时最短路径长度

解题代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define max(a , b) (a)>(b)?(a):(b)
#define min(a , b) (a)<(b)?(a):(b)
#define LL long long
using namespace std;

int v[500000];
int u[500000];
double w[500000];
int p[100];
int r[500000];
struct point
{
   int   x , y;
}points[60];
double dis(int i , int j )
{
    return  sqrt(pow(points[i].x-points[j].x,2) + pow(points[i].y-points[j].y,2));
}
int find(int x) 
{
   return  p[x] == x? x: p[x] = find(p[x]);
}
double cmp(const int i , const int j)
{
    return w[i] < w[j];
}
int main()
{
     int n;
    //freopen("/home/plac/problem/input.txt","r",stdin);

     while(scanf("%d",&n) != EOF)
     {
         memset(p,0,sizeof(p));
         memset(r,0,sizeof(r));    
         memset(v,0,sizeof(v));
         memset(w,0,sizeof(w));
         memset(u,0,sizeof(u));
         memset(points,0,sizeof(points));
         if(n == 0 )
            break;
         int a, b ; 
         scanf("%d %d",&a,&b);
         for(int i = 1;i <= n;i ++)        
         {
             scanf("%d %d",&points[i].x,&points[i].y);        
         }
         int k = 0;
         for(int i  = 1;i <= n;i ++)
          for(int j  = i+1;j <= n; j++)
         {
             v[k] = i ;
             u[k] = j ;
             w[k] = dis(i,j);
            // printf("%lf\n",w[k]);
             k ++;  
         }
         for(int i = 1;i <=  n;i ++) p[i] = i ; 
         for(int i = 0;i <= k ; i++) r[i]  = i ;
         sort(r,r+k,cmp);
         p[b] = a;
         double ans = dis(a,b);
         
         for(int i = 0 ;i < k ;i ++)
         {
             int e = r[i];
             int x  = find(v[e]);
             int y = find(u[e]);
             if(x != y)
             {
                ans += w[e];
                p[x] = y;
             }
         }          
         printf("%.2f\n",ans);
     }
    return 0 ;
}

 因为r数组开小了 ，导致溢出，，WA