Stall Reservations（贪心+优先队列）

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.

思路：

首先根据挤奶时间的先后顺序排序。。。然后将第一头牛加入优先队列。。然后就是加入优先队列的牛应该根据越早结束挤奶那么优先级更高，如果时间结束点相等，那么开始时间早的优先级高。。。

然后从前向后枚举。如果碰到有牛的挤奶时间的开始值大于优先队列的首部的结束值，那么说明这两头牛可以一起公用一个挤奶房。。然后从优先队列中删除这头牛。。那么这个问题就得到解决了。。。这道题用优先队列主要是优化时间复杂度；

#include<string.h>
#include<stdio.h>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
struct node
{
   int start_time;
   int end_time;
   int num;
    friend bool operator<(node a,node b)
   {
       if( a.end_time==b.end_time)
       return a.start_time>b.start_time;
       return a.end_time>b.end_time;
   }

}arr[50050],now;
bool cmp(node a,node b)
{
    if(a.start_time==b.start_time)
        return a.end_time<b.end_time;
    return a.start_time<b.start_time;
}
int ans[50050]={0};
int main()
{
    int i,n,l=1,MAX=-1;
    while(scanf("%d",&n)!=-1)
    {
        l=1;
        priority_queue<node>s,q;
    for(i=0;i<n;i++)
    {
        scanf("%d%d",&arr[i].start_time,&arr[i].end_time);
        arr[i].num=i;
    }
    sort(arr,arr+n,cmp);
    s.push(arr[0]);
    ans[s.top().num]=l;
    for(i=1;i<n;i++)
    {
            now=s.top();
          if(arr[i].start_time>now.end_time)
          {
              ans[arr[i].num]=ans[now.num];
              s.pop();
          }
          else
              ans[arr[i].num]=++l;
              s.push(arr[i]);
    }
      printf("%d\n",l);
     for(i=0;i<n;i++)
     {
         printf("%d\n",ans[i]);

     }
    }
    return 0;
}