Building Block(并查集优化)

最新推荐文章于 2024-12-24 15:04:34 发布

LBJHan

最新推荐文章于 2024-12-24 15:04:34 发布

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分类专栏：并查集

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本文介绍了一种构建块游戏中的算法实现，通过并查集数据结构处理两类操作：M操作将一个块移动到另一个块之上，C操作计算指定块下方的块数。详细解释了如何初始化并查集、查找根节点及进行合并操作。

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Building Block

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5939 Accepted Submission(s): 1842

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

Sample Output

1
0
2

题目说有N个板子，两个操作，M X Y表示把X板子所在的那一堆（X所在集合）放到Y板子所在的那一堆（Y所在集合）；C X表示求X板子下边有几个板子；、

直接上代码：

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#define MAX 30000+5
using namespace std;
int pre[MAX];//存父节点；
int d[MAX];//深度；
int r[MAX];//到根的距离；
int init(){ //初始化；
    for(int i=0; i<=MAX; i++){//从0开始，数据的一个坑点；
        pre[i]=i;
        d[i]=0;
        r[i]=1;
    }
}
int Find(int x){
    int fx;
    if(x==pre[x]) return x;
    fx=pre[x];
    pre[x]=Find(fx);
    d[x]=d[x]+d[fx];
    return pre[x];
}
void Union(int x, int y){
    int fx=Find(x), fy=Find(y);
    if(fx==fy) return;
    pre[fx]=fy;     //把x放到y上，计算x下边的block数量，则以y为根节点，所求值则变为x到根节点的距离；
    d[fx]=r[fy];
    r[fy]+=r[fx];
    return;
}
int main(){
    int P;
    scanf("%d",&P);
    init();
    while(P--){
        char op[5];
        int x, y;
        scanf("%s",op);
        if(op[0]=='M'){
            scanf("%d%d",&x,&y);
            Union(x,y);
        }
        else{
            scanf("%d",&x);
            Find(x);
            printf("%d\n",d[x]);
        }
    }
    return 0;
}