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NC24961. Hotel

描述

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 r to 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.

输入描述

* 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

输出描述

* 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.

示例1

输入:

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

输出:

1
4
7
0
5

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

C++14(g++5.4) 解法, 执行用时: 590ms, 内存消耗: 876K, 提交时间: 2019-07-27 15:44:53

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int room,t,num[100000];
int main()
{
	
	scanf("%d%d",&room,&t);
	for(int i = room;i > 0;i--)
	{
		num[i] = num[i+1]+1;
	}
	for(int i = 0;i < t;i++)
	{
		int temp;
		scanf("%d",&temp);
		if(temp == 1)
		{
			int nu,flag=1;
			scanf("%d",&nu);
			for(int i = 1;i <= room;i++)
			{
				if(num[i] >= nu)
				{
					flag = 0;
					printf("%d\n",i);
					for(int k = i;k < i + nu;k++)
					{
						num[k] = 0;
					}
					break;
				}
			}
			if(flag)puts("0");
			
		}
		else
		{
			int l,r;
			scanf("%d%d",&l,&r);
			for(int i = l + r - 1;i >= 1;i--)
			{
				if(i < l && num[i] == 0)break;
				num[i] = num[i + 1] + 1;
			}
			
		}
	}
}

C++11(clang++ 3.9) 解法, 执行用时: 730ms, 内存消耗: 868K, 提交时间: 2019-07-22 18:53:55

#include<stdio.h>
#define maxn 50005
int a[maxn];
int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	for(int i=n;i>=1;i--)
	{
		a[i]=a[i+1]+1;
	}
	while(m--)
	{
		int p;
		scanf("%d",&p);
		if(p==1)
		{
			int num,flag=1;
			scanf("%d",&num);
			for(int i=1;i<=n;i++)
			{
				if(a[i]>=num)
				{
					flag=0;
					printf("%d\n",i);
					for(int k=i;k<i+num;k++)
					{
						a[k]=0;
					}
					break;
				}
			}
			if(flag) printf("0\n");
		}
		if(p==2)
		{
			int l,r;
			scanf("%d %d",&l,&r);
			for(int i=l+r-1;i>=1;i--)
			{
				if(i<l&&a[i]==0) break;
				a[i]=a[i+1]+1;
			}
		}
	}
	return 0;
 } 

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