Stall Reservations

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10801		Accepted: 3819		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

5
1 10
2 4
3 6
5 8
4 7

Sample Output

4
1
2
3
2
4

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.

Source

USACO 2006 February Silver

题意：一群很有个性的奶牛，只在固定的时间产奶，每头牛需要用一个挤奶机器，问为满足所有牛产奶，最少需要多少个挤奶机器，并按照奶牛给出的顺序来输出该奶牛挤奶机器的编号。

 

思路： 创建结构体Cow，储存奶牛的编号、开始时间和完成时间。先对奶牛的开始时间排序，找到使用挤奶器的顺序。创建结构体Stall储存挤奶器的结束时间（实际上就是使用这个挤奶器的牛的完成时间）和挤奶器的编号（因为最后要输出使用挤奶器的个数和编号），采用优先队列的方法（挤奶器的结束时间为根据）保证结束时间早的可以被合适的重新利用。而优先队列中又有三种情况，第一，没有使用挤奶器，也就是没有奶牛；第二，有奶牛，但是结束不符合；第三种，有奶牛，结束时间也符合。

需要注意的是 这里边的变量很多，需要仔细的弄清楚他们的含义，以及对于优先队列的掌握情况。

这篇博客，讲的优先队列很透彻  https://blog.youkuaiyun.com/c20182030/article/details/70757660

话不多说，上代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
struct Cow{
	int a,b;
	int No;
	bool operator < (const Cow &c) const{
		return a<c.a;
	}
}cows[50100];
int pos[50100];
struct Stall{
	int end;
	int No;
	bool operator < (const Stall &s) const{
		return end>s.end;
	}
	Stall(int e,int n):end(e),No(n){}
};
int main(){
	int n;
	scanf("%d",&n);//
	for(int i=0;i<n;i++){
		scanf("%d %d",&cows[i].a,&cows[i].b);
		cows[i].No=i;///
	}
	sort(cows,cows+n);
	int total=0;
	priority_queue<Stall> pq;
	for(int i=0;i<n;i++){//
		if(pq.empty()){
			total++;
			pq.push(Stall(cows[i].b,total));
			pos[cows[i].No]=total;
		}
		else{
			Stall st=pq.top();
			if(st.end<cows[i].a){//
				pq.pop();
				pos[cows[i].No]=st.No;
				pq.push(Stall(cows[i].b,st.No));
			}
			else{
				total++;
				pq.push(Stall(cows[i].b,total));
				pos[cows[i].No]=total;	
			}
		}
	}
	printf("%d\n",total);
	for(int i=0;i<n;i++){
		printf("%d\n",pos[i]);
	}
	return 0;
}