【POJ - 3667】【Hotel】

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 Di (1 ≤ Di ≤ N) and approach the front desk to check in. Each group i requests a set of Di contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers r..r+Di-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 Xi and Di which specify the vacating of rooms Xi ..Xi +Di-1 (1 ≤ Xi ≤ N-Di+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 Di (b) Three space-separated integers representing a check-out: 2, Xi, and Di

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

10 6
1 3
1 3
1 3
1 3
2 5 5
1 6

Sample Output

1
4
7
0
5

需要处理好的是；两个子区间之间的关系，必须将这个考虑进去。

区间合并的题目，难就难在怎么将他们合并起来，这道题就是要记录一下每次入住多少人，问咱们再来人能不能住进去，如果能输出最小的那个房间号，如果不能 直接输出0，这道题在结构体里开了一个all 用来记录区间剩余的房间数目

ac代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define maxn 500005
using namespace std;

struct node{
	int lt;
	int rt;
	int left;
	int rigt;
	int laz;
	int all;
}tree[maxn];

void change(int cur,int val)
{
	tree[cur].left=(tree[cur].rt-tree[cur].lt+1)*val;
	tree[cur].rigt=(tree[cur].rt-tree[cur].lt+1)*val;
	tree[cur].all=(tree[cur].rt-tree[cur].lt+1)*val;
	return ; 
}
void pushup(int cur)
{
	tree[cur].left=tree[cur<<1].left;
	if(tree[cur<<1].left==(tree[cur<<1].rt-tree[cur<<1].lt+1))
		tree[cur].left+=tree[cur<<1|1].left;
	tree[cur].rigt=tree[cur<<1|1].rigt;
	if(tree[cur<<1|1].rigt==(tree[cur<<1|1].rt-tree[cur<<1|1].lt+1))
		tree[cur].rigt+=tree[cur<<1].rigt;
	tree[cur].all=max(tree[cur<<1].rigt+tree[cur<<1|1].left,max(tree[cur<<1].all,tree[cur<<1|1].all));
	return ;
}
void pushdown(int cur)
{
	if(tree[cur].laz!=-1)
	{
		change(cur<<1,tree[cur].laz);
		change(cur<<1|1,tree[cur].laz);
		tree[cur<<1].laz=tree[cur].laz;
		tree[cur<<1|1].laz=tree[cur].laz;
//		tree[cur<<1].all=tree[cur<<1].left=tree[cur<<1].rigt=tree[cur<<1].laz? 0:(m-(m>>1));
//		tree[cur<<1|1].all=tree[cur<<1|1].left=tree[cur<<1|1].rigt=tree[cur].laz?0:(m>>1);
		tree[cur].laz=-1;
	}
	return ;
}
void build(int l,int r,int cur)
{	
	tree[cur].lt=l;
	tree[cur].rt=r;
	tree[cur].laz=-1;
	if(l==r)
	{
		tree[cur].left=tree[cur].rigt=tree[cur].all=1;
		return ;
	}
	int m=(l+r)>>1;
	build(l,m,cur<<1);
	build(m+1,r,cur<<1|1);
	pushup(cur);
	return;
}

void updata(int l,int r,int val,int cur)
{
	if(l<=tree[cur].lt&&tree[cur].rt<=r)
	{
		change(cur,val);
		tree[cur].laz=val;
		return;
	}
	pushdown(cur);
	if(l<=tree[cur<<1].rt) updata(l,r,val,cur<<1);
	if(r>=tree[cur<<1|1].lt) updata(l,r,val,cur<<1|1);
	pushup(cur);
	return ;	
}
int  query(int  val,int cur)
{
	if(tree[cur].all<val) 
		return -1;
	if(tree[cur].lt==tree[cur].rt)
		return tree[cur].lt;
	pushdown(cur);
	if(tree[cur<<1].all>=val) 
		return query(val,cur<<1);
	else if(tree[cur<<1].rigt+tree[cur<<1|1].left>=val) 
		return tree[cur<<1].rt-tree[cur<<1].rigt+1;
	else 
		return query(val,cur<<1|1);
}


int main()
{
	int n,m;
	int op;
	int a,b;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		build(1,n,1);
		while(m--)
		{
			scanf("%d",&op);
			if(op==1)
			{
				scanf("%d",&a);
				int p=query(a,1);
				if(p!=-1)
				{
					updata(p,p+a-1,0,1);
					printf("%d\n",p);
				}
				else
					printf("0\n");
			}
			else if(op==2)
			{
				scanf("%d%d",&a,&b);
				updata(a,a+b-1,1,1);
			}
		}
	}
	return 0;
}