Stall Reservations -贪心+优先队列-区间问题

【题目】

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.

【输入】

Line 1: A single integer, N <br> <br>Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

【输出】

Line 1: The minimum number of stalls the barn must have. <br> <br>Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

【样例】

输入：

5
1 10
2 4
3 6
5 8
4 7

输出：

4
1
2
3
2
4

题目大意：一群奶牛，有格子固定的产奶时间，每头牛产奶时需要产奶机器，为了满足所有牛产奶，最后需要多少个产奶机器，并顺序输出他们需要的产奶机器的编号

思路：

1. 先按照牛产奶开始时间从小到大排序，如果开始时间相同，则按照结束时间从小到大排序

2. 用一个优先产奶结束时间的优先队列去维护当前产奶奶牛

3. 若下一个奶牛的开始产奶时间小于或等于当前奶牛的产奶结束时间，则需要一个新的产奶机器ans++，下一奶牛加入队列，若大于当前奶牛的产奶结束时间，则不需要新的产奶机器，并更新当前产奶奶牛

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int per[50010];
struct node
{
    int st,en,pos;

}cow[50010],now;

bool operator<(const node & a,const node & b)
{
    if(a.en==b.en)
        return a.st>b.st;
    return a.en>b.en;
}

int cmp(node a,node b)
{
    if(a.st==b.st)
        return a.en<b.en;
    return a.st<b.st;
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&cow[i].st,&cow[i].en);
            cow[i].pos=i;                          //数据初始编号
        }
        sort(cow,cow+n,cmp);
        memset(per,0,sizeof(per));
        priority_queue<node>q;
        q.push(cow[0]);
        per[cow[0].pos]=1;
        int ans=1;
        for(int i=1;i<=n-1;i++)
        {
            now=q.top();
            if(cow[i].st>now.en)
            {
                per[cow[i].pos]=per[now.pos];
                q.pop();
            }
            else
            {
                ans++;
                per[cow[i].pos]=ans;
            }
            q.push(cow[i]);
        }
        printf("%d\n",ans);
        for(int j=0;j<n;j++)
            printf("%d\n",per[j]);
    }
    return 0;
}