CodeForces 566D Restructuring Company 并查集的区间合并

题目链接：CodeForces 566D Restructuring Company



思路：
操作1：直接合并。

操作2：由于是一个区间段，区间所在集合编号确定。下次这个区间再有合并操作了，直接拿区间编号合并。所以用个数组记录，当前集合下一个集合的编号。

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>

using namespace std;
typedef long long ll;
const int MOD = 10000007;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int maxn = 200050;
int father[maxn];
int a[maxn];	//a[i]指向下一个与i不在同一集合里的元素

int find(int x)
{
	if(father[x]!=x)
		father[x] = find(father[x]);
		
	return father[x];
}

void Union(int x,int y)
{
	x = find(x);
	y = find(y);
	if(x!=y)
		father[y] = x;
}

bool judge(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x==y)
        return true;
    else 
        return false;
}


int main()
{
	int n,q;
	scanf("%d%d",&n,&q);
	
	for(int i=1;i<=n;i++)
	{
		father[i] = i;
		a[i] = i+1;
	}
		
	
	for(int i=0;i<q;i++)
	{
		int op,x,y;
		scanf("%d%d%d",&op,&x,&y);
		
		if(op==1)
		{
			Union(x,y);
		}
		else if(op==2)
		{
			int tmp;
			for(int i=x+1;i<=y;i=tmp)
			{
				Union(i-1,i);
				tmp = a[i];
				a[i] = y+1; 	//该段区间所有元素的下一个元素都是y+1
			}
		}
		else
		{
			if(judge(x,y))
				printf("YES\n");
			else
				printf("NO\n");
		}
	}

	
	return 0;
}