Hotel 旅馆 题解(From luoguBlog)

考试前深陷分块泥潭所以刚开始以为是分块。

然而这题数据水到暴力卡常都能AC

正解：万物皆可线段树

节点存储区间长度、区间最长连续空房长度、从左往右最长连续空房长度、从右往左最长连续空房长度。

维护后三个值即可。

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int n,m;
const int N=50010;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct node
{
	int maxl,lm,rm,lazy,len;
	#define mid (l+r>>1)
	#define ls (k<<1)
	#define rs (k<<1|1)
	#define maxl(k) tree[k].maxl
	#define lm(k) tree[k].lm
	#define rm(k) tree[k].rm
	#define lazy(k) tree[k].lazy
	#define len(k) tree[k].len
}tree[N<<2];
void up(int k)
{
    lm(k)=(lm(ls)==len(ls))?len(ls)+lm(rs):lm(ls);
    rm(k)=(rm(rs)==len(rs))?len(rs)+rm(ls):rm(rs);
    maxl(k)=max(rm(ls)+lm(rs),max(maxl(ls),maxl(rs)));
}
void downlazy(int k)
{
    if(!lazy(k)) return ;
    lazy(ls)=lazy(rs)=lazy(k);
    maxl(ls)=lm(ls)=rm(ls)=(lazy(k)==1)?len(ls):0;
    maxl(rs)=lm(rs)=rm(rs)=(lazy(k)==1)?len(rs):0;
    lazy(k)=0;
}
void build(int l,int r,int k)
{
    maxl(k)=lm(k)=rm(k)=len(k)=r-l+1;
    if(l==r) return ;
    build(l,mid,ls);
    build(mid+1,r,rs);
}
void update(int L,int R,int op,int l,int r,int k)
{
    if(L <= l && r <= R){
        maxl(k)=lm(k)=rm(k)=(op==1)?len(k):0;
        lazy(k)=op;
        return ;
    }
    downlazy(k);
    if(L<=mid) update(L,R,op,l,mid,ls);
    if(R>mid) update(L,R,op,mid+1,r,rs);
    up(k);
}
int query(int len,int l,int r,int k)
{
    if(l==r) return l;
    downlazy(k);
    if(maxl(ls)>=len) return query(len,l,mid,ls);
    if(rm(ls)+lm(rs)>=len) return mid-rm(ls)+1;
    return query(len,mid+1,r,rs);
}
int main()
{
	n=read();m=read();
	build(1,n,1);
	int op,x,d;
	while(m--)
	{
		op=read();
		if(op==1)
		{
			d=read();
			if(maxl(1)>=d)
			{
				x=query(d,1,n,1);
				printf("%d
",x);
				update(x,x+d-1,2,1,n,1);
			}
			else puts("0");
		}
		else
		{
			x=read();d=read();
			update(x,x+d-1,1,1,n,1);
		}
	}
	return 0;
}