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\n",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;
}