POJ 3667 Hotel

最新推荐文章于 2018-05-30 19:51:52 发布

原创最新推荐文章于 2018-05-30 19:51:52 发布 · 766 阅读

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本文介绍了一个用于处理酒店房间分配请求的高效算法。该算法通过维护一个特殊的数状数组来跟踪房间占用状态，并能够快速响应入住和退房操作，确保客户能够获得连续的可用房间。

Hotel

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 9597		Accepted: 4113

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 r to 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

num数组记录该区间是否住慢，0代表都是空的，1代表都住了人。-1代表有的住了，有的空着。

d[rt][0]代表从区间最左端开始的空的最长长度，d[rt][1]代表从区间最右端开始的空的最长长度，d[rt][2]代表区间的最长空的长度。

#include<cstdio>

#define maxn  51111
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Max(a,b) (a>b?a:b)

int num[maxn<<2],d[maxn<<2][3];

void PushDown(int rt,int tmp)
{
	if(num[rt]==-1)
		return ;
	num[rt<<1]=num[rt<<1|1]=num[rt];
	if(num[rt]==0)
	{
		d[rt<<1][0]=d[rt<<1][1]=d[rt<<1][2]=(tmp+1)>>1;
		d[rt<<1|1][0]=d[rt<<1|1][1]=d[rt<<1|1][2]=tmp>>1;
	}
	else
	{
		d[rt<<1][0]=d[rt<<1][1]=d[rt<<1][2]=0;
		d[rt<<1|1][0]=d[rt<<1|1][1]=d[rt<<1|1][2]=0;
	}
	num[rt]=-1;
}

void build(int l,int r,int rt)
{
	num[rt]=0;
	for(int i=0;i<3;i++)
		d[rt][i]=r-l+1;
	if(l==r)
		return ;
	int m=(l+r)>>1;
	build(lson);
	build(rson);
}

int check(int l,int r,int rt,int D)								//查找满足条件的最左点
{
	if(d[rt][2]<D)
		return 0;
	PushDown(rt,r-l+1);
	int m=(l+r)>>1;
	if(d[rt<<1][2]>=D)
		return check(lson,D);
	if(d[rt<<1][1]+d[rt<<1|1][0]>=D)
		return m-d[rt<<1][1]+1;
	return check(rson,D);
}

void PushUp(int rt,int tmp)
{
	if(d[rt<<1][0] < (tmp+1)>>1)
		d[rt][0]=d[rt<<1][0];
	else
		d[rt][0]=((tmp+1)>>1)+d[rt<<1|1][0];

	if(d[rt<<1|1][1]<tmp>>1)
		d[rt][1]=d[rt<<1|1][1];
	else
		d[rt][1]=d[rt<<1][1]+(tmp>>1);

	d[rt][2]=Max(d[rt<<1][2],d[rt<<1|1][2]);
	d[rt][2]=Max(d[rt][2],d[rt<<1][1]+d[rt<<1|1][0]);
}

void updata(int l,int r,int rt,int L,int R,int flag)
{
	if(l>=L && r<=R)
	{
		num[rt]=flag;
		if(flag==1)
			d[rt][0]=d[rt][1]=d[rt][2]=0;
		else
			d[rt][0]=d[rt][1]=d[rt][2]=r-l+1;
		return ;
	}
	PushDown(rt,r-l+1);
	int m=(l+r)>>1;
	if(L<=m)
		updata(lson,L,R,flag);
	if(R>m)
		updata(rson,L,R,flag);
	PushUp(rt,r-l+1);
}

int main()
{
	int N,M;
	int flag,X,D,tmp;

	while(scanf("%d%d",&N,&M)==2)
	{
		build(1,N,1);

		while(M--)
		{
			scanf("%d",&flag);

			if(flag==1)
			{
				scanf("%d",&D);
				tmp=check(1,N,1,D);
				printf("%d\n",tmp);
				if(tmp!=0)
				{
					updata(1,N,1,tmp,tmp+D-1,1);
				}
			}
			if(flag==2)
			{
				scanf("%d%d",&X,&D);
				updata(1,N,1,X,X+D-1,0);
			}
		}
	}

	return 0;
}