P2894 [USACO08FEB]Hotel G

https://www.luogu.com.cn/problem/P2894

线段树区间连续最大的1的个数，加上区间修改，lazy覆盖。虽然不难，写出来还是很interesting的。。

具体看代码吧

#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 1e5+11;
int ans[4*maxn],L[4*maxn],R[4*maxn];

int lazy[maxn*4];

void up(int node,int be,int en){
	int l = node*2;
	int r = node*2+1;
	int mid = be + en  >> 1;
	
	ans[node] = max(ans[l],ans[r]);
	ans[node] = max(ans[node],L[r] +  R[l]);
	
	L[node] = L[l]; 
	if(mid - be + 1 == L[l]) L[node] = mid - be + 1 + L[r];
	
	R[node] = R[r];
	if(en - mid == R[r]) R[node] = en - mid + R[l];
}



int push(int node,int be,int en){
	int mid = be + en >> 1;
	int l = node * 2;
	int r = node * 2 + 1;
	if(lazy[node]){
		ans[l] = L[l] = R[l] = (mid - be + 1)*(lazy[node]-1);
		ans[r] = L[r] = R[r] = (en - mid)*(lazy[node]-1);
		lazy[l] = lazy[r] = lazy[node];
		lazy[node] = 0;
	}
	return 0;
}
int update(int node,int be,int en,int LL,int RR,int val){//直接用1覆盖 
	int mid = be + en >> 1;
	int l = node*2;
	int r = node*2 + 1;
	
	if(LL <= be && en <= RR){
		lazy[node] = val;
		ans[node] = L[node] = R[node] = (en - be + 1)*(val-1);
		return 0;
	}
	
	push(node,be,en);
	if(LL <= mid) update(l,be,mid,LL,RR,val);
	if(RR > mid) update(r,mid+1,en,LL,RR,val);
	up(node,be,en);
	return 0;
}

int id = 0;

int ask(int node,int be,int en,int x){
	int mid = be + en >> 1;
	int l = node*2;
	int r = node*2 + 1;
	if(be == en){
		id = be;
		return 0;
	}
	
	push(node,be,en);
	if(ans[l] >= x){
		ask(l,be,mid,x);
		up(node,be,en);
	}
	else if(R[l] +  L[r] >= x){
		id = mid - R[l] + 1;
		return 0;
	}
	else if(ans[r] >= x){
		ask(r,mid+1,en,x);
		up(node,be,en);
	}
	else{
		return 0;
	}
	return 0;
}

int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	update(1,1,n,1,n,2);
	id= 0 ;
	while(m--){
		int op;
		scanf("%d",&op);
		if(op == 1){
			int x;
			scanf("%d",&x);
			id = 0;
			ask(1,1,n,x);
			if(id != 0) update(1,1,n,id,id+x-1,1);
			printf("%lld
",id);
		}
		else{
			int x,y;
			scanf("%d %d",&x,&y);
			update(1,1,n,x,x + y - 1,2);
		}
	}
	return 0;
}