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[];

 #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[];

 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 = ; i < n; i++)
         {
             scanf("%d %d",&cows[i].a,&cows[i].b);
             cows[i].c = i;
         }
         sort(cows,cows+n,cmp1);
         int  ans = ;
         cows[].flag = ans;
         que.push(cows[]);

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