POJ 3667 Hotel （线段树区间合并）

最新推荐文章于 2019-09-21 01:33:36 发布

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#线段树 #合并 #poj #iostream

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本文介绍了一种使用线段树解决区间合并问题的方法，通过具体的酒店房间分配与回收案例进行阐述。该方法能够高效地处理大量连续区间的空间分配与回收请求。

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Language:

Hotel

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 12417		Accepted: 5346

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

Source

USACO 2008 February Gold

题意：一个旅馆有n件连续的房子，开始都为空，有m个操作，每一个操作，如果第一个数为1 ，下一个数为le,那么输出长度le的最左边的编号，并且把这le个房间住人（如果没有那么多空房输出0），如果第一个数为2 后面 le,ri,那么把le 到le+ri-1的房间清空

就是线段树区间合并，需要注意的一个问题我在代码中标出了：

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
using namespace std;

#define N 50005

int n,m;

struct stud{
int le,ri;
int ls,rs,ms;
int cover;
}f[N*4];

void pushdown(int pos)
{
	if(f[pos].cover==-1) return ;

	f[L(pos)].ls=f[L(pos)].rs=f[L(pos)].ms=f[pos].cover ?0:f[L(pos)].ri-f[L(pos)].le+1;
	f[R(pos)].ls=f[R(pos)].rs=f[R(pos)].ms=f[pos].cover ?0:f[R(pos)].ri-f[R(pos)].le+1;

    f[L(pos)].cover=f[R(pos)].cover=f[pos].cover;
    f[pos].cover=-1;
    //唯一需要注意的一点就是把f[pos].cover更改为-1，因为如果不改，下次还会pushdown
//这里它已经把值pushdown了，下次不需要pushdown

}

void pushup(int pos)
{
	f[pos].ls=f[L(pos)].ls;
	f[pos].rs=f[R(pos)].rs;
	f[pos].ms=max(f[L(pos)].ms,f[R(pos)].ms);

	if(f[L(pos)].ls==f[L(pos)].ri-f[L(pos)].le+1)
		f[pos].ls+=f[R(pos)].ls;

	if(f[R(pos)].rs==f[R(pos)].ri-f[R(pos)].le+1)
		f[pos].rs+=f[L(pos)].rs;

	f[pos].ms=max(f[pos].ms,f[L(pos)].rs+f[R(pos)].ls);
}

void build(int pos,int le,int ri)
{
	f[pos].le=le;
	f[pos].ri=ri;
	f[pos].ls=f[pos].rs=f[pos].ms=ri-le+1;
	f[pos].cover=0;
	if(le==ri) return ;

	int mid=MID(le,ri);
	build(L(pos),le,mid);
	build(R(pos),mid+1,ri);

	pushup(pos);
}

void update(int pos,int le,int ri,int va)
{
	if(f[pos].le==le&&f[pos].ri==ri)
	{
		f[pos].ls=f[pos].rs=f[pos].ms=va? 0:ri-le+1;
		f[pos].cover=va;
		return ;
	}

    pushdown(pos);

    int mid=MID(f[pos].le,f[pos].ri);

    if(mid>=ri)
		update(L(pos),le,ri,va);
	else
		if(mid<le)
		update(R(pos),le,ri,va);
	else
	{
		update(L(pos),le,mid,va);
		update(R(pos),mid+1,ri,va);
	}

  pushup(pos);
}

int query(int pos,int le)
{
	if(f[pos].le==f[pos].ri)
		return f[pos].le;

	pushdown(pos);

	int mid=MID(f[pos].le,f[pos].ri);

	if(f[L(pos)].ms>=le)
		return query(L(pos),le);
	else
		if(f[L(pos)].rs+f[R(pos)].ls>=le)
		return mid-f[L(pos)].rs+1;

	return query(R(pos),le);

}

int main()
{
	int i,j,x,le,ri;
	while(~scanf("%d%d",&n,&m))
	{

		build(1,1,n);

		while(m--)
		{
			scanf("%d",&x);
			if(x==1)
			{
				scanf("%d",&le);
				if(f[1].ms<le)
				{
					printf("0\n");
				}
				else
				{
					int ans=query(1,le);
					printf("%d\n",ans);
					update(1,ans,ans+le-1,1);
				}
			}
           else
		   {
		   	scanf("%d%d",&le,&ri);
		   	update(1,le,le+ri-1,0);
		   }
		}
	}
    return 0;
}