UVA11987--Almost Union-Find--带权并查集

最新推荐文章于 2024-12-05 09:43:38 发布

SCUT_Pein

最新推荐文章于 2024-12-05 09:43:38 发布

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分类专栏：数据结构_并查集文章标签： Almost Union-Find UVA11987

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本文介绍了一种改进的并查集数据结构实现方法，该方法支持三种操作：合并两个集合、从一个集合中移除元素及查询指定集合的元素数量和元素总和。通过对并查集进行特殊处理，使得在支持元素删除的同时保持高效的性能。

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Description

I hope you know the beautiful Union-Find structure. In this problem, you're to implement something similar, but not identical.

The data structure you need to write is also a collection of disjoint sets, supporting 3 operations:

1 p q : Union the sets containing p and q. If p and q are already in the same set, ignore this command.

2 p q : Move p to the set containing q. If p and q are already in the same set, ignore this command

3 p : Return the number of elements and the sum of elements in the set containing p.

Initially, the collection contains n sets: {1}, {2}, {3}, ..., {n}.

Input Description

There are several test cases. Each test case begins with a line containing two integers n and m (1<=n,m<=100,000), the number of integers, and the number of commands. Each of the next m lines contains a command. For every operation, 1<=p,q<=n. The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.

Output Description

For each type-3 command, output 2 integers: the number of elements and the sum of elements.

Sample Input

Sample Output

3 12
3 7
2 8

/*
这是一个需要删除的并查集。我尝试过用father和son。比较麻烦。而且最终超时了。。
看了题解，需要删除点的并查集，转换为带权并查集，直接将需要删除的点的权值置0；
操作1：将两个并查集合并，得同时合并num和sum.
操作2：比如删除x.idx[x]表示真正的x在哪里。先将此个x权值置0.再开个新的点。然后合并该点和q即可
*/
#include <iostream>
#include <cstdio>
using namespace std;
#define maxn 200018
int father[maxn],idx[maxn],num[maxn];//num来多少个，idx才存
long long int sum[maxn];
int n,m,cnt;
int find(int x)
{
	if(x==father[x])return x;
	return find(father[x]);
}
void init()
{
	for(int i=1;i<=n;i++)
	{
		father[i]=idx[i]=sum[i]=i;
		num[i]=1;
	}
	cnt=n;
}
void Union(int p,int q)
{
	int pp=find(idx[p]),qq=find(idx[q]);
	father[pp]=qq;
	num[qq]+=num[pp];
	sum[qq]+=sum[pp];
}
void Delete(int p)
{
	int pp=idx[p];
	sum[find(pp)]-=p;
	num[find(pp)]--;
	idx[p]=++cnt;sum[idx[p]]=p;num[idx[p]]=1;father[idx[p]]=idx[p];
}
int main()
{
	while(scanf("%d%d",&n,&m)==2)
	{
		init();
		int ope,p,q;
		for(int i=1;i<=m;i++)
		{
			scanf("%d",&ope);
			if(ope==1)
			{
				scanf("%d%d",&p,&q);
				if(find(idx[p])==find(idx[q]))continue;
				else Union(p,q);
			}
			else if(ope==2)
			{
				scanf("%d%d",&p,&q);
				if(find(idx[p])!=find(idx[q]))
				{
					Delete(p);
					Union(p,q);
				}
			}
			else 
			{
				int u;
				scanf("%d",&u);
				int fuck=find(idx[u]);
				printf("%d %lld\n",num[fuck],sum[fuck]);//这里我用I64d既然WA。。。
			}
		}
	}
	return 0;
}