poj1988

并查集 r[]记录节点到跟节点的距离 num[] 树的节点数

#include <stdio.h>
int father[30005],r[30005],num[30001];
int find(int x)
{
    if(x!=father[x])
    {
        int t = father[x];
        father[x] = find(father[x]);
        r[x]+=r[t];
    }
    return father[x];
}

void union1(int x, int y)
{
    father[y] = x;
    r[y]+=num[x];
    num[x]+=num[y] ;

}
int main()
{
    int n, i, j, x, y, k;
    char c;
    scanf("%d", &n);
    for(i = 1 ; i <= 30000 ; i++)
    {
        father[i] = i;
        r[i] = 0;
        num[i] = 1;
    }
    while(n--)
    {
        getchar();
        scanf("%c", &c);
        if(c == 'M')
        {
            scanf("%d%d", &x,&y);
            int pa = find(x);
            int pb = find(y);
            if(pa!=pb)
            {
                union1(pa,pb);
            }
        }
        else
        {
            scanf("%d", &x);
            y = find(x);
            printf("%d\n", num[y]-r[x]-1);
        }
    }
    return 0;
}

#include <stdio.h>
 int father[30005],r[30005],num[30001];
 int find(int x)
 {
     if(x!=father[x])
     {
         int t = father[x];//这个跟下面的那条不能反 保证t不为跟节点
         father[x] = find(father[x]);//回溯时把节点都指向根节点  方便下次查找     
         r[x]+=r[t];
     }
     return father[x];
 }

 void union1(int x, int y)
 {
     father[y] = x;
     r[y]+=num[x];
     num[x]+=num[y] ;

 }
 int main()
 {
     int n, i, j, x, y, k;
     char c;
     scanf("%d", &n);
     for(i = 1 ; i <= 30000 ; i++)
     {
         father[i] = i;
         r[i] = 0;
         num[i] = 1;
     }
     while(n--)
     {
         getchar();
         scanf("%c", &c);
         if(c == 'M')
         {
             scanf("%d%d", &x,&y);
             int pa = find(x);
             int pb = find(y);
             if(pa!=pb)
             {
                 union1(pa,pb);
             }
         }
         else
         {
             scanf("%d", &x);
             y = find(x);
             printf("%d\n", num[y]-r[x]-1);
         }
     }
     return 0;
 }