Building Block

                                                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768
K (Java/Others)

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

Source

2009 Multi-University Training Contest 1 - Host
by TJU

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点击打开题目链接

有N个砖头，编号为1～N。有两种操作， 第一种是M  x  y,  把 x 所在的那一堆砖头全部移动放到 y 所在的那堆上面。  第二种操作是C  x，  即查询 x 下面有多少个砖头。

带权并查集的应用， 用 Num[ x ] 来表示以 x 为根的这棵树的节点总数(x 必须是根，没有父亲节点)， Under[ x ] 数组来表示 x 砖头下面有多少个砖头。

如果把a堆砖放到b堆砖上面， 那么a堆最底面那个砖头root_a的下面原本是有0个砖头的， 搬过去之后，变成了有 b 堆砖的数量个。然后root_a之上的数量便根据root_a的值进行更新。更新的步骤在查找时路径压缩的那一步进行。

#include <cstdio>
#include <iostream>
using namespace std;

const int MAXN = 30000 + 10;
int Fa[MAXN], Under[MAXN], Num[MAXN];

void Init()
{
    for (int i = 0; i < MAXN; i++)
    {
        Fa[i] = i;              //初始化并查集，自成一个连通分量
        Num[i] = 1;             //i是根节点，初始化以它为根的树共有1个节点
    }
}

int Find(int x)
{
    if (x == Fa[x]) return x;
    int fa = Fa[x];
    Fa[x] = Find(Fa[x]);        //递归查找老祖宗
    Under[x] += Under[fa];      //加上父亲的
    return Fa[x];
}

void Union(int x, int y)
{
    int tx = Find(x), ty = Find(y); //先查找两者的祖先
    if (tx == ty) return;
    Fa[tx] = ty;                //合并
    Under[tx] = Num[ty];        //tx下面多了Num[ty]个
    Num[ty] += Num[tx];         //以ty为根的树多了Num[tx]个节点
    Num[tx] = 0;                //以tx为根的树没有结点（此时tx已经不是一个树的真正根节点）
}

int main()
{
    int t, x, y;
    char cmd[10];
    Init();
    scanf("%d", &t);
    while (t--)
    {
        scanf("%s", cmd);
        if (cmd[0] == 'C')
        {
            scanf("%d", &x);
            Find(x);
            printf("%d\n", Under[x]);
        }
        else
        {
            scanf("%d%d", &x, &y);
            Union(x, y);
        }
    }
    return 0;
}