Stall Reservations - POJ3190 贪心

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本文介绍了一个关于牛棚分配的问题，通过算法确定最少需要多少个牛棚以确保每头牛能在指定的时间内产奶且不发生冲突。使用了优先队列来高效地管理牛棚的分配与回收。

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3541		Accepted: 1266		Special Judge

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.

Sample Input

Sample Output

Hint

Explanation of the sample:

Here's a graphical schedule for this output:

Time     1  2  3  4  5  6  7  8  9 10

Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>

Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..

Stall 3 .. .. c3>>>>>>>>> .. .. .. ..

Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible.

题意：每个牛有一个特定的产奶时间，如果两个牛的产奶时间冲突的话就要新加一个牛棚，问最少需要多少个牛棚让每头牛都能产奶并且不互相冲突，输出最小数目及每头牛所在的牛棚编号

思路：由于每头牛的产奶时间是固定的，所以对其排序后再遍历时只需看结束时间最早的那个牛棚能否允许当前的牛前去产奶，能的话更新牛棚的结束时间，不能的话申请新的牛棚，利用优先队列维护牛棚

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#define MAXN 50100
using namespace std;
struct node
{
	int s,e;
	int num,id;
}cow[MAXN];
bool cmp(node a,node b)
{
	return a.s<b.s;
}
bool cmp1(node a,node b)
{
	return a.num<b.num;
}
struct stall
{
	int T,ID;
	stall(int t,int n){
		T=t;ID=n;
	}
	bool operator< (const stall &b) const {
		return T>b.T;
	}
};
priority_queue<stall> qq;
int main()
{
//	freopen("in.txt","r",stdin);
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d%d",&cow[i].s,&cow[i].e);
		cow[i].num=i;
	}
	sort(cow+1,cow+1+n,cmp);
	for(int i=1;i<=n;i++) {
		if(!qq.empty()){
			stall temp=qq.top();
			if(cow[i].s>temp.T){
				temp.T=cow[i].e;
				cow[i].id=temp.ID;
				qq.pop();
				qq.push(temp);
			}else {
				cow[i].id=qq.size()+1;
				qq.push(stall(cow[i].e,cow[i].id));
			}
		}else{
			cow[i].id=1;
			qq.push(stall(cow[i].e,1));
		}
	}
	printf("%d\n",qq.size());
	sort(cow+1,cow+1+n,cmp1);
	for(int i=1;i<=n;i++)
		printf("%d\n",cow[i].id);
}