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.
 
 
贪心思想，先把输入的时间段升序排序，创建一个优先队列，优先知道最靠左时间段，方便共用stall。
 
代码：
 
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdio>
using namespace std;

class time
{
    public:
    int a,b;
    int id;
    friend bool operator <(time x,time y)
    {
        if(x.b==y.b)return x.a>y.a;///右端点相同 则取左端点靠前的 即最先被分配的
        return x.b>y.b;///堆顶元素，右端点最小
    }
}s[50001];
int cmp(time x,time y)
{
    if(x.a==y.a)return x.b < y.b;///左端点相同的 右端点也升序
    return x.a < y.a;///左端点升序排序
}
int main()
{
    int n,used[50001],ans = 0;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&s[i].a,&s[i].b);
        s[i].id=i;
    }
    sort(s,s+n,cmp);
    priority_queue<time> q;

    for(int i = 0;i < n;i ++)
    {
        if(!q.empty()&&s[i].a > q.top().b)///和队顶比较  队顶保存的是最靠前的，如果满足条件可以插在它的后面，公用一个stall
        {
            used[s[i].id]=used[q.top().id];
            q.pop();
        }
        else///否则另加一个stall
        {
            ans++;
            used[s[i].id]=ans;
        }
        q.push(s[i]);
    }
    cout<<ans<<endl;
    for(int i=0;i<n;i++)
        cout<<used[i]<<endl;
}
View Code