POJ 3190 Stall Reservations - 贪心+优先队列

最新推荐文章于 2022-03-20 21:20:29 发布

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最新推荐文章于 2022-03-20 21:20:29 发布

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分类专栏：贪心

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本文介绍了一种解决奶牛挤奶时间冲突问题的算法。该算法使用优先队列来实现最小化所需挤奶位数量的目标，并确保每头奶牛能在指定时间内单独挤奶。文章提供了完整的代码实现及解析。

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Stall Reservations

Description

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows.

Help FJ by determining:

The minimum number of stalls required in the barn so that each cow can have her private milking period
An assignment of cows to these stalls over time

Many answers are correct for each test dataset; a program will grade your answer.

Input

Line 1: A single integer, N

Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

Output

Line 1: The minimum number of stalls the barn must have.

Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

考虑到应该需要排列然后贪心，但是每次判断的项因为每一个数的插入二改变，这时可以考虑利用优先队列，注意此题应用优先队列为最小堆，需要在struct里面进行友元定义

至于priority_queue的简单的大小顶堆定义也可以通过声明时来定义:

priority_queue<TYPE,vector<TYPE>,greater<TYPE>> Q;

其中greater<TYPE>为大顶堆，less<TYPE>为小顶堆

#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn=50010;

int s[maxn],ans,n;
struct point
{
	int a,b,cnt;
	friend bool operator < (point x,point y)
	{
		return x.b>y.b;
	}
}p[maxn];
priority_queue<point> q;

bool cmp(point x,point y)
{
	return x.a<y.a||(x.a==y.a&&x.b<y.b);
}
int main()
{
	while(~scanf("%d",&n))
	{
		ans=0;
		memset(s,0,sizeof(s));
		memset(p,0,sizeof(p));
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d",&p[i].a,&p[i].b);
			p[i].cnt=i;
		}
		sort(p+1,p+n+1,cmp);
		s[p[1].cnt]=++ans; //N.B.此处并非s[1]，而应是排列后的第一项
		q.push(p[1]);
		for(int i=2;i<=n;i++)
		{
			point tmp=q.top();
			if(tmp.b<p[i].a)
			{
				s[p[i].cnt]=s[tmp.cnt];
				q.pop();
			}
			else 
				s[p[i].cnt]=++ans;
			q.push(p[i]);
		}
		printf("%d\n",ans);
		for(int i=1;i<=n;i++)
			printf("%d\n",s[i]);
	}
	return 0;
}