POJ 3667 Hotel

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

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

Sample Output

1
4
7
0
5

线段树区间修改+谜之查询。
想到还是比较简单的。
re的同学可以试一下 1 1 1 1
很神奇哦！

#include<iostream>
#include<cstdio>
using namespace std;
const int N=100005;
int n,m;
struct node
{
	int l,r,smx,lmx,rmx,tag;
}a[4*N];
void build(int num,int l,int r)
{
	a[num].l=l,a[num].r=r;
	a[num].tag=-1;
	if(l==r)
	{
		a[num].smx=a[num].lmx=a[num].rmx=1;
		return ;
	}
	int mid=(l+r)/2;
	build(2*num,l,mid),build(2*num+1,mid+1,r);
	a[num].rmx=a[num].lmx=a[num].smx=a[2*num].smx+a[2*num+1].smx;
}
int query(int num,int d)
{
	if(a[2*num].smx>=d)
		return query(2*num,d);
	else if(a[2*num].rmx+a[2*num+1].lmx>=d)
		return a[2*num].r-a[2*num].rmx+1;
	else if(a[2*num+1].smx>=d)
		return query(2*num+1,d);
	else
		return 1;
}
void check(int num)
{
	if(a[num].l!=a[num].r)
	{
		if(a[num].tag==1)
		{
			a[2*num].rmx=a[2*num].lmx=a[2*num].smx=0;
			a[2*num+1].rmx=a[2*num+1].lmx=a[2*num+1].smx=0;
		}
		else
		{
			a[2*num].rmx=a[2*num].lmx=a[2*num].smx=a[2*num].r-a[2*num].l+1;
			a[2*num+1].rmx=a[2*num+1].lmx=a[2*num+1].smx=a[2*num+1].r-a[2*num+1].l+1;
		}
		a[2*num].tag=a[2*num+1].tag=a[num].tag;
	}
	a[num].tag=-1;
}
void change(int num,int l,int r,int x)
{
	if(a[num].l>=l&&a[num].r<=r)
	{
		if(x==1)
			a[num].rmx=a[num].lmx=a[num].smx=0;
		else
			a[num].rmx=a[num].lmx=a[num].smx=a[num].r-a[num].l+1;
		a[num].tag=x;
		return ;
	}
	if(a[num].l>r||a[num].r<l)
		return ;
	if(a[num].tag!=-1)
		check(num);
	change(2*num,l,r,x);
	change(2*num+1,l,r,x);
	a[num].lmx=a[2*num].lmx,a[num].rmx=a[2*num+1].rmx;
	a[num].smx=max(max(a[2*num].smx,a[2*num+1].smx),a[2*num].rmx+a[2*num+1].lmx);
	if(a[2*num].lmx==a[2*num].r-a[2*num].l+1)
		a[num].lmx=a[2*num].lmx+a[2*num+1].lmx;
	if(a[2*num+1].rmx==a[2*num+1].r-a[2*num+1].l+1)
		a[num].rmx=a[2*num].rmx+a[2*num+1].rmx;
}
int main()
{
	scanf("%d%d",&n,&m);
	build(1,1,n);
	while(m--)
	{
		int d,k,x;
		scanf("%d",&k);
		if(k==1)
		{
			scanf("%d",&d);
			if(a[1].smx<d)
				printf("0\n");
			else
			{
				x=query(1,d);
				printf("%d\n",x);
				change(1,x,x+d-1,1);
			}
		}
		if(k==2)
		{
			scanf("%d%d",&x,&d);
			change(1,x,x+d-1,0);
		}
	}
	return 0;
}