Educational Codeforces Round 23 F. MEX Queries 离散化+线段树

F. MEX Queries

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 l r — Add all missing numbers from the interval [l, r]
• 2 l r — Remove all present numbers from the interval [l, r]
• 3 l 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 ≤ 105).

Next n lines contain three integer numbers t, l, r (1 ≤ t ≤ 3, 1 ≤ l ≤ r ≤ 1018) — type of the query, left and right bounds.

Output

Print MEX of the set after each query.

Examples

input

3
1 3 4
3 1 6
2 1 3

output

1
3
1

input

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

output

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

题意：给你n个区间，1操作表示把[l,r]赋值为1，2操作表示把[l,r]赋值为0，3操作表示把[l,r]异或1；

　　   求第一个不为0的正整数。

思路：对于所有区间[l,r]取出l，r，r+1三个点；

　　　离散化放入线段树中去

　　　现在需要区间修改，区间异或，查询

　　　需要两个lazy标记，一个存修改的值，一个存是否异或1.

　　　区间查询log查找即可，详见代码。

　　　小trick：需要加入最小值，即是1的情况；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<bitset>
#include<time.h>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=3e5+,M=4e6+,inf=,mod=1e9+;
const LL INF=1e18+,MOD=1e9+;

int L=3e5+;
struct LT
{
    int sum[N<<],to[N<<],sw[N<<];
    void pushdown(int pos,int l,int r)
    {
        int mid=(l+r)>>;
        if(to[pos]!=-)
        {
            to[pos<<]=to[pos];
            to[pos<<|]=to[pos];
            sum[pos<<]=to[pos]*(mid-l+);
            sum[pos<<|]=to[pos]*(r-mid);
            to[pos]=-;
            sw[pos]=;
        }
        if(sw[pos])
        {
            if(to[pos<<]!=-)to[pos<<]=!to[pos<<];
            else sw[pos<<]=!sw[pos<<];
            if(to[pos<<|]!=-)to[pos<<|]=!to[pos<<|];
            else sw[pos<<|]=!sw[pos<<|];
            sum[pos<<]=(mid-l+)-sum[pos<<];
            sum[pos<<|]=(r-mid)-sum[pos<<|];
            sw[pos]=;
        }
    }
    void build(int l,int r,int pos)
    {
        to[pos]=-;
        sw[pos]=;
        sum[pos]=;
        if(l==r)return;
        int mid=(l+r)>>;
        build(l,mid,pos<<);
        build(mid+,r,pos<<|);
    }
    void update(int L,int R,int c,int l,int r,int pos)
    {
        if(L<=l&&r<=R)
        {
            if(c==)
            {
                if(to[pos]!=-)to[pos]=!to[pos];
                else sw[pos]=!sw[pos];
            }
            else
            {
                to[pos]=c;
                sw[pos]=;
            }
            if(c==)sum[pos]=r-l+-sum[pos];
            else sum[pos]=(r-l+)*c;
            return;
        }
        pushdown(pos,l,r);
        int mid=(l+r)>>;
        if(L<=mid)update(L,R,c,l,mid,pos<<);
        if(R>mid) update(L,R,c,mid+,r,pos<<|);
        sum[pos]=sum[pos<<|]+sum[pos<<];
    }
    int query(int l,int r,int pos)
    {
        //cout<<l<<" "<<r<<" "<<sum[pos]<<" "<<to[pos]<<endl;
        if(l==r)return l;
        pushdown(pos,l,r);
        //cout<<sum[pos<<1]<<" "<<sum[pos<<1|1]<<endl;
        int mid=(l+r)>>;
        if(sum[pos<<]==mid-l+)return query(mid+,r,pos<<|);
        else return query(l,mid,pos<<);
    }
}tree;

int t[N],len;
LL l[N],r[N],s[N];
int getpos(LL x)
{
    int pos=lower_bound(s+,s+len,x)-s;
    return pos;
}
int main()
{
    int n,k=;
    scanf("%d",&n);
    s[++k]=;
    for(int i=;i<=n;i++)
        scanf("%d%lld%lld",&t[i],&l[i],&r[i]),s[++k]=l[i],s[++k]=r[i],s[++k]=r[i]+;
    sort(s+,s++k);
    len=unique(s+,s++k)-s;
    tree.build(,L,);
    for(int i=;i<=n;i++)
    {
        int z=getpos(l[i]),y=getpos(r[i]);
        if(t[i]==)tree.update(z,y,,,L,);
        else if(t[i]==)tree.update(z,y,,,L,);
        else if(t[i]==)tree.update(z,y,,,L,);
        int x=tree.query(,L,);
        //cout<<"xxx "<<x<<" "<<tree.sum[1]<<endl;
        printf("%lld\n",s[x]);
    }
    return ;
}