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
      
       #include
       
        #include
        
         #include
         
          #include
          
           #define max(a , b) (a)>(b)?(a):(b)#define min(a , b) (a)<(b)?(a):(b)#define LL long longusing 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