codeforces CF817F MEX Queries 线段树 离散化

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F. MEX Queries

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time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output

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You are given a set of integer numbers, initially it is empty. You should perform $ n $ queries.
There are three different types of queries:
　

• $ 1 quad l quad r $ — Add all missing numbers from the interval $ [l, r] $
• $ 2 quad l quad r $ — Remove all present numbers from the interval $ [l, r] $
• $ 3 quad l quad r $ — Invert the interval $ [l, r] $ — add all missing and remove all present numbers from the interval $ [l, r] $
  　

After each query you should output $ MEX $ of the set
— the smallest positive $ (MEX  ≥ 1) $ integer number which is not presented in the set.
　

Input

The first line contains one integer number $ n (1 ≤ n ≤ 10^5) $ .
Next $ n $ lines contain three integer numbers $ t, l, r (1 ≤ t ≤ 3, 1 ≤ l ≤ r ≤ 10^18) $ — type of the query, left and right bounds.
　

Output

Print $ MEX $ of the set after each query.
　

Examples

input1

 3
 1 3 4
 3 1 6
 2 1 3

output1

 1 
 3
 1

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input2

 4
 1 1 3
 3 5 6
 2 4 4
 3 1 6

output2

 4
 4
 4
 1

　

Note

Here are contents of the set after each query in the first example:

1. $ (3, 4) $ — the interval $ [3, 4] $ is added
2. $ (1, 2, 5, 6) $ — numbers $ (3, 4) $ from the interval $ [1, 6] $ got deleted and all the others are added
3. $ (5, 6) $ — numbers $ (1, 2) $ got deleted
   　

题目大意

• 给你一个无限长的数组，初始的时候都为 $ 0 $ ，
  操作 $ 1 $ 是把给定区间设为 $ 1 $ ，
  操作 $ 2 $ 是把给定区间清零，
  操作 $ 3 $ 是把给定区间反转。
  每次操作后要输出最小位置的 $ 0 $ 。

• 数据范围：$ 1 le n le 10^5 $
  　

思路

• 线段树+离散化

• 操作 $ 1 $ 和 $ 2 $ ：区间赋值

• 操作 $ 3 $ ：相当于区间每个数 $ xor quad 1 $

• 可以维护线段树的区间长度，线段树上二分即可

• 注意：

1. 求的是 $ mex $ ，可能不是 $ l_i,r_i $ 中的任何一个，对于 $ 1,l_{i+1},r_{i+1} $ 也要离散。
2. 记录原来信息的数组必须得开 $ long quad long $
   　

代码

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#define ll long long
#define N 400005
inline ll read() {
    register char ch;
    while(!isdigit(ch=getchar()));
    register ll x=ch^'0';
    while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
    return x;
}
int n,m,opt[N];
ll a[N<<2],ql[N],qr[N],sum[N<<2],lzy[N<<2],rev[N<<2];
inline void pushdown(int o,int l,int r){
	int mid=l+r>>1;
	if(lzy[o]){
		lzy[o<<1]=lzy[o<<1|1]=lzy[o]--;
		sum[o<<1]=(mid-l+1)*lzy[o];
		sum[o<<1|1]=(r-mid)*lzy[o];
		lzy[o]=rev[o<<1]=rev[o<<1|1]=0;
	}
	if(rev[o]){
		sum[o<<1]=mid-l+1-sum[o<<1];
		sum[o<<1|1]=r-mid-sum[o<<1|1];
		rev[o<<1]^=1;
		rev[o<<1|1]^=1;
		rev[o]=0;
	}
}
void updata(int o,int l,int r,int L,int R,int val){
	if(L<=l&&r<=R){
		lzy[o]=val--; rev[o]=0;
		sum[o]=(val)*(r-l+1);
		return;
	}
	pushdown(o,l,r);
	int mid=l+r>>1;
	if(L>mid) updata(o<<1|1,mid+1,r,L,R,val);
	else if(R<=mid) updata(o<<1,l,mid,L,R,val);
	else{
		updata(o<<1,l,mid,L,R,val);
		updata(o<<1|1,mid+1,r,L,R,val);
	}
	sum[o]=sum[o<<1]+sum[o<<1|1];
}
void modify(int o,int l,int r,int L,int R){
	if(L<=l&&r<=R){
		sum[o]=r-l+1-sum[o];
		rev[o]^=1;
		return;
	}
	pushdown(o,l,r);
	int mid=l+r>>1;
	if(L>mid) modify(o<<1|1,mid+1,r,L,R);
	else if(R<=mid) modify(o<<1,l,mid,L,R);
	else{
		modify(o<<1,l,mid,L,R);
		modify(o<<1|1,mid+1,r,L,R);
	}
	sum[o]=sum[o<<1]+sum[o<<1|1];	
}
ll query(int o,int l,int r){
	if(l==r) return a[l];
	pushdown(o,l,r);
	int mid=l+r>>1;
	ll res;
	if(sum[o<<1]!=mid-l+1) res=query(o<<1,l,mid);
	else res=query(o<<1|1,mid+1,r);
	sum[o]=sum[o<<1]+sum[o<<1|1];
	return res;
}
signed main(){
	n=read(); a[++m]=1;
	for(int i=1;i<=n;++i){
		opt[i]=read(); ql[i]=read(); qr[i]=read();
		a[++m]=ql[i]; a[++m]=ql[i]+1;
		a[++m]=qr[i]; a[++m]=qr[i]+1;
	}
	sort(a+1,a+1+m);
	m=unique(a+1,a+1+m)-(a+1);
	for(int i=1;i<=n;++i){
		int L=lower_bound(a+1,a+1+m,ql[i])-a;
		int R=lower_bound(a+1,a+1+m,qr[i])-a;
		if(opt[i]==1||opt[i]==2) updata(1,1,m,L,R,opt[i]==1 ? 2 : 1);
		else modify(1,1,m,L,R);
		printf("%lld
",query(1,1,m));
	}
	return 0;
}
/*
#        42702002
When     2018-09-09 14:03:14 
Who      PotremZ 
Problem  F - MEX Queries 
Lang     GNU C++11
Verdict  Accepted 
Time     264 ms 
Memory   57900 KB 
*/