poj3667-Hotel 线段树区间合并

最新推荐文章于 2019-08-15 11:59:25 发布

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最新推荐文章于 2019-08-15 11:59:25 发布

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分类专栏： poj随笔线段树文章标签：线段树 poj

本文链接：https://blog.youkuaiyun.com/wang_heng199/article/details/75322246

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本文介绍了一种使用线段树解决酒店房间预订问题的方法，通过区间合并思想及延迟标记来高效处理连续房间的入住与退房请求。

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Hotel

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 17772		Accepted: 7707

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, andD_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

题意：给你旅馆数量，现在可以进去一些人，还可以退出一些人，这些人都是连续的。

解题思路：使用了线段树的区间合并思想，需要使用延迟标记，在更新过程中。

ac代码：
#include <cstdio> 
#include <iostream>
using namespace std;
#define Max 50010
struct node{
	int left, right, maxn;
	int co;//延迟标记 
}sum[Max<<2];

int max(int a, int b)
{
	return a>b?a:b;
}

void Build(int l, int r, int rt)
{
	sum[rt].co = -1;
	sum[rt].left = sum[rt].right = sum[rt].maxn = r - l +1;
   if(l == r)	 return ;
   int m = (l + r) >> 1;
   Build(l, m, rt<<1);
   Build(m+1, r, rt<<1|1);
}

void Pushdown(int rt, int k)
{
  if(sum[rt].co != -1){
  	sum[rt<<1].co = sum[rt<<1|1].co = sum[rt].co;
  	sum[rt<<1].left = sum[rt<<1].right = sum[rt<<1].maxn = sum[rt].co ? 0 : (k-(k>>1));
  	sum[rt<<1|1].left = sum[rt<<1|1].right = sum[rt<<1|1].maxn = sum[rt].co ? 0 : (k>>1);

  	sum[rt].co = -1;
  }	
}

void Pushup(int rt, int k)//区间合并 
{
        sum[rt].left = sum[rt<<1].left ;	
        sum[rt].right = sum[rt<<1|1].right;
        if(sum[rt<<1].left == k-(k>>1)) sum[rt].left += sum[rt<<1|1].left;
        if(sum[rt<<1|1].right == k>>1) sum[rt].right += sum[rt<<1].right;  	
		sum[rt].maxn = max(max(sum[rt<<1].maxn, sum[rt<<1|1].maxn), sum[rt<<1].right + sum[rt<<1|1].left);
}

void update(int L, int R, int l, int r, int t, int rt)
{
   if(L <= l && R >= r)	{
   	sum[rt].left = sum[rt].right = sum[rt].maxn = t ? 0 : r-l+1;
   	sum[rt].co = t;
   	return ;
   }
   Pushdown(rt, r-l+1);
   int m = (l + r) >> 1;
   if(L <= m) update(L, R, l, m, t, rt<<1);
   if(m < R) update(L, R, m+1, r, t, rt<<1|1);
   Pushup(rt , r-l+1);
}

int query(int l, int r, int t, int rt)
{
	if(l == r) return 1;
	Pushdown(rt, r-l+1);
	int m = (l + r) >> 1;
	if(t <= sum[rt<<1].maxn) return query(l, m, t, rt<<1);
	else if(t <= sum[rt<<1].right + sum[rt<<1|1].left) return m - sum[rt<<1].right + 1;
	else return query(m+1, r, t, rt<<1|1);
}   

int main()
{
  int n, m;
  int p, a, b;
  while(~scanf("%d%d", &n, &m))
  {
  	Build(1, n, 1);
  while(m--)
  {
  	scanf("%d", &p);
  	if(p == 1){
  		scanf("%d", &a);
  		if(sum[1].maxn < a) printf("0\n");
  		else {
  			int pos = query(1, n, a, 1);
  			printf("%d\n", pos);
  			update(pos, pos+a-1, 1, n, 1, 1);
		  }
		  }
  		  else {
  		  	scanf("%d%d", &a, &b);
  		  	update(a, a+b-1, 1, n, 0, 1);
			}
  }
  }
  return 0;
}

题目链接：点击打开链接http://poj.org/problem?id=3667