poj 1988 Cube Stacking

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 23540		Accepted: 8247
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

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX = 30005;
int parent[MAX];
int sum[MAX];//若parent[i]=i,sum[i]表示砖块i所在堆的砖块数
int under[MAX];//under[i]表示砖块i下面有多少砖块

void init(){
    for(int i = 0; i < MAX; i++){
        parent[i] = i;
        sum[i] = 1;
        under[i] = 0;
    }
}

int GetParent(int a){//获取a的根，并把a的父节点改为根
    if(parent[a] == a)
        return a;
    int p = GetParent(parent[a]);
    under[a] += under[parent[a]];
    parent[a] = p;
    return parent[a];
}

void merge(int a, int b){
    //把b所在的堆，叠放到a所在的堆。
    int pa = GetParent(a);
    int pb = GetParent(b);
    if(pa == pb)
        return ;
    parent[pb] = pa;
    under[pb] = sum[pa];//under[pb]赋值前一定是0，因为parent[pb] = pb,pb一定是原b所在堆最底下的
    sum[pa] += sum[pb];
}

int main(){
    int p;
    init();
    scanf("%d", &p);
    for(int i = 0; i < p; i++){
        char s[20];
        int a, b;
        scanf("%s", s);
        if(s[0] == 'M'){
            scanf("%d%d", &a, &b);
            merge(b, a);
        }
        else {
            scanf("%d", &a);
            GetParent(a);
            printf("%d\n", under[a]);
        }
    }        
    return 0;
}