poj3667 Hotel

最新推荐文章于 2021-02-01 18:09:19 发布

Kirito_Acmer

最新推荐文章于 2021-02-01 18:09:19 发布

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分类专栏：线段树文章标签：线段树

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本文深入探讨了AI音视频处理领域中的关键技术，特别是视频分割与语义识别。通过详细解释这些技术的工作原理、应用案例及实际效果，旨在为读者提供全面的理解和洞察。此外，文章还涉及了这些技术如何应用于自动驾驶、AR增强现实等场景，展示其在现代技术发展中的重要作用。

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Description

The cows are journeying north to Thunder Bay in Canada to gain cultural enrichment and enjoy a vacation on the sunny shores of Lake Superior. Bessie, ever the competent travel agent, has named the Bullmoose Hotel on famed Cumberland Street as their vacation residence. This immense hotel has N (1 ≤ N ≤ 50,000) rooms all located on the same side of an extremely long hallway (all the better to see the lake, of course).

The cows and other visitors arrive in groups of size D_i (1 ≤ D_i ≤ N) and approach the front desk to check in. Each group i requests a set of D_i contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers r..r+D_i-1 if they are available or, if no contiguous set of rooms is available, politely suggests alternate lodging. Canmuu always chooses the value of rto be the smallest possible.

Visitors also depart the hotel from groups of contiguous rooms. Checkout i has the parameters X_i and D_i which specify the vacating of rooms X_i ..X_i +D_i-1 (1 ≤ X_i ≤ N-D_i+1). Some (or all) of those rooms might be empty before the checkout.

Your job is to assist Canmuu by processing M (1 ≤ M < 50,000) checkin/checkout requests. The hotel is initially unoccupied.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Line i+1 contains request expressed as one of two possible formats: (a) Two space separated integers representing a check-in request: 1 and D_i(b) Three space-separated integers representing a check-out: 2, X_i, and D_i

Output

* Lines 1.....: For each check-in request, output a single line with a single integer r, the first room in the contiguous sequence of rooms to be occupied. If the request cannot be satisfied, output 0.

Sample Input

Sample Output

1
4
7
0
5
看了别人的思路自己写出来了，感觉真棒！（笑）
这道题属于线段树区间更新，首先维护线段树6个变量l,r,llen,rlen,tlen,mark(l,r表示区间最左和最右的坐标，llen表示从区间的最左端开始向右连续有空位的总个数，rlen表示从区间的最右端开始向左连续有空位的总个数，tlen表示这个区间最大连续空位的个数，mark表示这段区间是否有空位的情况，1表示都占满了，0表示都是空位，-1表示既有空位又有被占的。这里为了提高线段树的时间效率采用了成段更新，即每次不更新到叶节点，当b[i].l==l && b[i].r==r时返回。
每段的tlen:是左右子区间的tlen以及左子树rlen加上右子树llen的最大值。
每段的rlen:是左子树的rlen,如果左子树都是空位，那么再加上右子树llen.
每段的llen:是右子树的llen,如果右子树都是空位，那么再加上左子树llen.
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<string>
using namespace std;
#define maxn 50006
int a[maxn];
struct node{
	int l,r,llen,rlen,tlen,mark;
}b[4*maxn];

void build(int l,int r,int i)
{
	int mid;
	b[i].l=l;b[i].r=r;b[i].mark=0;
	b[i].llen=b[i].rlen=b[i].tlen=r-l+1;
	if(l==r) return;
	mid=(l+r)/2;
	build(l,mid,i*2);
	build(mid+1,r,i*2+1);
}

void update(int l,int r,int mark,int i)
{
	int mid;
	if(b[i].mark==mark)return;
	if(b[i].l==l && b[i].r==r){
		b[i].mark=mark;
		if(mark==0)b[i].llen=b[i].rlen=b[i].tlen=r-l+1;
		else b[i].llen=b[i].rlen=b[i].tlen=0;return;
	}
	if(b[i].mark!=-1){
		b[i*2].mark=b[i*2+1].mark=b[i].mark;
		if(b[i].mark==0){
			b[i*2].llen=b[i*2].rlen=b[i*2].tlen=b[i*2].r-b[i*2].l+1;
			b[i*2+1].llen=b[i*2+1].rlen=b[i*2+1].tlen=b[i*2+1].r-b[i*2+1].l+1;
		}
		else if(b[i].mark==1){
			b[i*2].llen=b[i*2].rlen=b[i*2].tlen=0;
			b[i*2+1].llen=b[i*2+1].rlen=b[i*2+1].tlen=0;
		}
		b[i].mark=-1;
	}
	mid=(b[i].l+b[i].r)/2;
	if(r<=mid)update(l,r,mark,i*2);
	else if(l>mid)update(l,r,mark,i*2+1);
	else{
		update(l,mid,mark,i*2);
		update(mid+1,r,mark,i*2+1);
	}
	int temp=max(b[i*2].tlen,b[i*2+1].tlen);
	b[i].tlen=max(temp,b[i*2].rlen+b[i*2+1].llen);
	b[i].llen=b[i*2].llen;
	if(b[i*2].llen==b[i*2].r-b[i*2].l+1) b[i].llen+=b[i*2+1].llen;
	b[i].rlen=b[i*2+1].rlen;
	if(b[i*2+1].rlen==b[i*2+1].r-b[i*2+1].l+1) b[i].rlen+=b[i*2].rlen;
}

int question(int num,int i)
{
	int mid;
	if(b[i].tlen<num)return 0;
	if(b[i].l==b[i].r && num==1)return b[i].l;
	if(b[i].mark!=-1){
		b[i*2].mark=b[i*2+1].mark=b[i].mark;
		if(b[i].mark==0){
			b[i*2].llen=b[i*2].rlen=b[i*2].tlen=b[i*2].r-b[i*2].l+1;
			b[i*2+1].llen=b[i*2+1].rlen=b[i*2+1].tlen=b[i*2+1].r-b[i*2+1].l+1;
		}
		else if(b[i].mark==1){
			b[i*2].llen=b[i*2].rlen=b[i*2].tlen=0;
			b[i*2+1].llen=b[i*2+1].rlen=b[i*2+1].tlen=0;
		}
	}
	if(b[i].llen>=num)return b[i].l;
	else if(b[i*2].tlen>=num) return question(num,i*2);
	else if(b[i*2].rlen+b[i*2+1].llen>=num)return b[i*2].r-b[i*2].rlen+1;
	else return question(num,i*2+1);
}

int main()
{
	int n,m,i,j,a,b,c,ans;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		build(1,n,1);
		for(i=1;i<=m;i++){
			scanf("%d",&a);
			if(a==1){
				scanf("%d",&b);
				ans=question(b,1);
				printf("%d\n",ans);
				if(ans)update(ans,ans+b-1,1,1);
			}
			else{
				scanf("%d%d",&b,&c);
				update(b,b+c-1,0,1);
			}
		}
	}
	return 0;
}