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.

 

题解：整理区间，将不重合的区间整理成一个区间

需要用到优先队列

当优先队列的元素是结构体时候，需要在结构体内重载比较操作符函数。

struct node{
    int a;
    int b;
    int flag;
     friend bool operator <(node node1,node node2)  
    {  
        //<为从大到小排列，>为从小到大排列  
        return node1.b<node2.b;  
    }  
}n[5];

#include<iostream>
#include<algorithm> 
#include<queue>
using namespace std;

struct node{
    int a; //开始时间 
    int b; //结束时间 
    int c; //序列号 
    int flag;    //区间安排序号 
    friend bool operator <(node node1,node node2) //重载比较操作符函数 
    {
        return node1.b > node2.b;
    }
}cows[50005];

bool cmp1(node t1,node t2)  //根据区间排序 
{
    if(t1.a != t2.a) return t1.a < t2.a;
    else return t1.b < t2.b;
}

bool cmp2(node t1,node t2) //根据序列号排序 
{
     return t1.c < t2.c;
}


int main()
{
    int n;
    priority_queue<node>que;
    while(~scanf("%d",&n))
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%d %d",&cows[i].a,&cows[i].b);
            cows[i].c = i;
        }
        sort(cows,cows+n,cmp1);  
        int  ans = 1;
        cows[0].flag = ans;
        que.push(cows[0]);
        
        for(int i = 1 ; i < n; i++)
        {
            if(que.top().b >= cows[i].a)  //此时cows的区间与que内的任意区间有重合 
            {
                ans++;
                cows[i].flag = ans;
            }
            else
            {
                cows[i].flag = que.top().flag;
                que.pop();
            }
            que.push(cows[i]);
        }

        sort(cows,cows+n,cmp2);  
        cout<<ans<<endl;
        for(int i = 0; i < n; i++)
        {
            printf("%d\n",cows[i].flag);
        }
    }
    return 0;
}