带权并查集- Building Block （HDU - 2818）

最新推荐文章于 2020-06-02 09:56:13 发布

Nothing_227

最新推荐文章于 2020-06-02 09:56:13 发布

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分类专栏：并查集文章标签：带权并查集 HDU 2818

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本文介绍了一种处理块堆叠问题的算法实现，通过并查集数据结构进行高效的操作，包括将一个块堆叠到另一个块之上以及计算指定块下方的块数量。

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

题意：存在两种操作，M 操作和 C 操作；

1.M a b 操作：将a所在的块（就是a所在的那一堆）全部堆到 b（所在的那一堆上）。

2.C a 操作：计算a下面有多少个块。

由样例可知操作：M 1 6，M 2 4后的得到如图且根设置为 6 和 4；故而 fa[1] = 6,fa[2] = 4; cnt[4] = 2,cnt[6] = 2;

再经过 M 2 6后可知

此时的4号下面是原先的 6号集合故num[4]的值为cnt[6],fa[4] = 6.

所以当查找 2 号下的所有点时，递归路线为 2 -----> 4 -----> 6;

num[2] += (num[4]+=num[6]);

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 3e4+5;
int fa[maxn],num[maxn],cnt[maxn]; //cnt记录每个集合的个数，num记录下面块的个数
int findfa(int x)
{
    if(x == fa[x])
        return x;
    int temp = findfa(fa[x]); 
    num[x] += num[ fa[x] ]; //递归时不断的更新经过点的值
    return fa[x] = temp;
}
void Union(int a,int b)
{
    int x = findfa(a),y = findfa(b);
    if(x != y)
    {
        fa[x] = y; 
        num[x] = cnt[y]; //该跟结点下全是下面的节点
        cnt[y] += cnt[x]; //统计该集合所有点的个数
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    char c;
    int a,b;
    for(int i = 0; i < maxn; i++)
    {
        fa[i] = i;
        cnt[i] = 1;
    }
    memset(num,0,sizeof num);
    while(T--)
    {
        cin>>c;
        if(c == 'C')
        {
            scanf("%d",&a);
            findfa(a);
            cout<<num[a]<<endl;
        }
        else
        {
            scanf("%d%d",&a,&b);
            Union(a,b);
        }
    }
    return 0;
}
/*
6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4

*/