2 seconds
256 megabytes
standard input
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.
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.
Print MEX of the set after each query.
3
1 3 4
3 1 6
2 1 3
1
3
1
4
1 1 3
3 5 6
2 4 4
3 1 6
4
4
4
1
Here are contents of the set after each query in the first example:
- {3, 4} — the interval [3, 4] is added
- {1, 2, 5, 6} — numbers {3, 4} from the interval [1, 6] got deleted and all the others are added
- {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 ;
}