POJ 3190

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.

Line 1: A single integer, N

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

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.

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：n头牛，每头牛的挤奶时间，问你最少需要几台机器工作，然后对每头牛所工作机器的编号。

做法：贪心，排序，如果l相等，r小的排前面，否则l小的排前面，维护一个优先队列，正在工作的机器且优先结束的机器排在最前面，判断这台机器奶牛是否能用，如果不能用，那么再加一台机器。

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
struct node{
	int x,y,i;
	bool operator <(const node &a)const   {  
	    if(y==a.y)return x>a.x;  
	    return y>a.y;  //先结束的排前面
    }  
}e[50005];
bool cmp(node a,node b){
	if(a.x==b.y)return a.y<b.y;
	return a.x<b.x;
}
int use[50005];
priority_queue<node>sp;
int main(){
	int n,i,j;
	while(scanf("%d",&n)!=EOF){
		for(i=1;i<=n;i++){
			scanf("%d%d",&e[i].x,&e[i].y);
			e[i].i=i;
		}
		sort(e+1,e+1+n,cmp);
		int cnt=1;
		use[e[1].i]=1;
		sp.push(e[1]);
		for(i=2;i<=n;i++){
			if(!sp.empty()&&sp.top().y<e[i].x){//如果队列非空且最先结束的机器的时间小于当前奶牛开始时间
				use[e[i].i]=use[sp.top().i];
				sp.pop();
			}
			else{//否则就再加一台机器
				cnt++;
				use[e[i].i]=cnt;
			}
			sp.push(e[i]);
		}
		printf("%d\n",cnt);
		for(i=1;i<=n;i++){
			printf("%d\n",use[i]);
		}
		while(!sp.empty())sp.pop();
	}
	return 0;
}