POJ

Stall Reservations

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.       //！！！

 

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。。。

开始时间升序排序，从左往右排，不会出现排在队首左边的情况。（贪心）

总结：开始时间升序排，每个条件都用上，求含不重叠子序列的最少序列数；

           结束时间升序排，只用部分条件，求一个序列含不重叠子序列最多数。

#include<stdio.h>
#include<algorithm>
#include<queue>
using namespace std;

struct Node{
    int x,y,no;
    friend bool operator<(Node a,Node b)
    {
        if(a.y==b.y) return a.x>b.x;
        return a.y>b.y;
    }
}node[50005];
bool cmp(Node a,Node b)
{
    if(a.x==b.x) return a.y<b.y;
    return a.x<b.x;
}
int a[50005];
priority_queue<Node> q;
int main()
{
    int n,c,i;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%d%d",&node[i].x,&node[i].y);
        node[i].no=i;
    }
    sort(node+1,node+n+1,cmp);
    q.push(node[1]);
    a[node[1].no]=1;
    c=1;
    for(i=2;i<=n;i++){
        if(q.size()&&node[i].x>q.top().y){
            a[node[i].no]=a[q.top().no];
            q.pop(); 
        }
        else{
            c++;
            a[node[i].no]=c;
        }
        q.push(node[i]);
    }
    printf("%d
",c);
    for(i=1;i<=n;i++){
        printf("%d
",a[i]);
    }
    return 0;
}