Stall Reservations(贪心优先队列)

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10792		Accepted: 3814		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.

Source

USACO 2006 February Silver

1.学会优先队列的使用。

2.解题思路：

1) 把所有奶牛按开始时间从小到大排序。
2) 为第一头奶牛分配个畜栏。
3) 依次处理后面每头奶牛 i。处理 i时，考虑已分配畜栏中结束间最早的畜栏 x。
若 E(x) < S( E(x) < S( E(x) < S( i), 则不用分配新畜栏， i可进入 x, 并修改 E(x) 为E( i)
若 E(x) >= S( E(x) >= S( E(x) >= S(E(x) >= S( i)，则分配新畜栏 y, 记 E( y) = E( y) = i)
直到所有奶牛处理结束
始终需要用优先队列存放已经分配的畜栏，并使得结束时间最早始终需要用优先队列存放已经分配的畜栏。

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

struct cows
{
	int s;
	int end;
	int no;
	bool operator<(const cows &c)const
	{
		return s<c.s;
	}
}cow[50006]; 
int pos[50006];

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;
	int all=0;
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>cow[i].s>>cow[i].end;
		cow[i].no=i;
	}
	sort(cow,cow+n);
	priority_queue<stall>q;
	for(int i=0;i<n;i++)
	{
		if(q.empty())//没有空的，直接申请 
		{
			all++;//栅栏数加一 
			q.push(stall(cow[i].end,all));//将该牛结束的时间作为栅栏开始使用的时间，all作为编号 
			pos[cow[i].no]=all;
		}
		else
		{
			stall sta=q.top();//优先队列 (大根堆模型)
			if(sta.end<cow[i].s)//如果栅栏结束的时间 ，比下一头牛申请的开始时间小，就可以使用 
			{
				q.pop();
				pos[cow[i].no]=sta.no;
				q.push(stall(cow[i].end,sta.no));
			}
			else//都不合适，还得重新申请 
			{
				all++;
				q.push(stall(cow[i].end,all));
				pos[cow[i].no]=all;
			}
		}
	}
	cout<<all<<endl;
	for(int i=0;i<n;i++)
	{
		cout<<pos[i]<<endl;
	}
	return 0;
}

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