hdu 2818--Building Block

Problem Description

John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 <= P <= 1000000). There are two kinds of operation:

M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command. 
C X : Count the number of blocks under block X 

You are request to find out the output for each C operation.

 

 

Input

The first line contains integer P. Then P lines follow, each of which contain an operation describe above.

 

 

Output

Output the count for each C operations in one line.

 

 

Sample Input

6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4

 

 

Sample Output

1 0 2

 

这题算是一个简单的带权并查集的应用，问题中包含了两种操作，M操作，将包含x的block全部移到y上去，C操作询问在x下的block数是多少

如果直接存储x下的block数会很麻烦，我们还不如求x上面的blocks总数，最后用它父节点所有的子节点数-x上面的blocks数-x本身就得到了

要求的。对于带权并查集的应用，表示才刚开始学，所了解的也只有简单的几种。貌似这题跟黑书上的那个差不多。。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=30010;
int father[N],rank[N],cnt[N];
void init()
{
    for(int i=0;i<=N-10;i++)
    {
        father[i]=i;
        rank[i]=1;
        cnt[i]=0;
    }
}
int find(int x)
{
    if(x==father[x])
        return x;
    int t=father[x];
    father[x]=find(father[x]);
    cnt[x]+=cnt[t];
    return father[x];
}
void merge(int x,int y)
{
    x=find(x),y=find(y);
    if(x==y)
        return ;
    father[y]=x;
    cnt[y]=rank[x];
    rank[x]+=rank[y];
}
int main()
{
    int p,u,v;
    char cmd[2];
    while(scanf("%d",&p)!=EOF)
    {
        init();
        while(p--)
        {
            scanf("%s",cmd);
            if(cmd[0]=='M')
            {
                scanf("%d%d",&u,&v);
                merge(u,v);
            }
            else
            {
                scanf("%d",&u);
                v=find(u);
                printf("%d\n",rank[v]-cnt[u]-1);
            }
        }
    }
    return 0;
}