POJ 1988 Cube Stacking（并查集）

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 15900		Accepted: 5419
Case Time Limit: 1000MS

Description

Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations: 
moves and counts. 
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y. 
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value. 

Write a program that can verify the results of the game. 

Input

* Line 1: A single integer, P 

* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X. 

Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself. 

Output

Print the output from each of the count operations in the same order as the input file. 

Sample Input

6

M 1 6

C 1

M 2 4

M 2 6

C 3

C 4

Sample Output

1

0

2

Source

USACO 2004 U S Open

 

 

 

/*

POJ 1988

*/

#include <stdio.h>

#include <iostream>

#include <algorithm>

#include <string.h>

using namespace std;

const int MAXN=30010;

int F[MAXN];

int num[MAXN];//堆的数目

int val[MAXN];//下面的个数

int find(int x)

{

    if(F[x]==-1)return x;

    int tmp=find(F[x]);

    val[x]+=val[F[x]];

    return F[x]=tmp;

}

void bing(int u,int v)//将u所在的堆放在v上面，注意顺序

{

    int t1=find(u),t2=find(v);

    if(t1!=t2)

    {

        F[t1]=t2;

        val[t1]=num[t2];

        num[t2]+=num[t1];

    }

}

int main()

{

    int P;

    int u,v;

    char op[10];

    while(scanf("%d",&P)==1)

    {

        for(int i=0;i<MAXN;i++)

        {

            F[i]=-1;

            val[i]=0;

            num[i]=1;

        }

        while(P--)

        {

            scanf("%s",&op);

            if(op[0]=='C')

            {

                scanf("%d",&u);

                find(u);

                printf("%d\n",val[u]);

            }

            else

            {

                scanf("%d%d",&u,&v);

                bing(u,v);

            }

        }

    }

    return 0;

}