HDU 2818 Building Block （带权并查集）

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本文介绍了一种使用带权并查集解决石子堆叠问题的方法，通过维护每个石子的高度和其下方石子数量来高效处理石子堆叠与查询操作。

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Building Block

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

Total Submission(s): 167 Accepted Submission(s): 74

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

Source

2009 Multi-University Training Contest 1 - Host by TJU

题意一排石子， M a b 代表把a石子放到b上， C a 代表查询 a石子下面有多少石子。

用带权并查集，我们把最下面的点当作根节点，high[]表示当前石子的高度， below[]表示当前石子下面有多少石子

#include <iostream>
#include <cmath>
#include <string>
#include <algorithm>
#include <string.h>
using namespace std;
const int MAXN = 30005;
int pre[MAXN], below[MAXN], high[MAXN];
void init() {
    for (int i = 0; i < MAXN; ++i) {
        pre[i] = i;
        below[i] = 0;//自己下面的
        high[i] = 1;//高度
    }
}
int find(int v) {
        //cout << v << "   " << pre[v] << "   " << below[pre[v]] << endl;
    if (pre[v] == v)
        return v;
    //回溯更新父节点，以致更新当前节点
    int temp = pre[v];
    pre[v] = find(pre[v]);
    below[v] += below[temp];
        //cout << v << "   " << temp << "   " << below[temp] << endl;
    return pre[v];
}
int main() {
    int n;
    char ch[2];
    while (cin >> n) {
        init();
        for (int i = 0; i < n; ++i) {
            cin >> ch;
            if (ch[0] == 'M') {
                int a, b;
                cin >> a >> b;
                int fa = find(a), fb = find(b);
                if (fa == fb)
                    continue;
                pre[fa] = fb;
                below[fa] = high[fb];
                high[fb] += high[fa];
            } else if (ch[0] == 'C') {
                int a;
                cin >> a;
                find(a);
                cout << below[a] << endl;
            }
        }
    }
}