（3）A - 贪心

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.

题意：给出每个奶牛的挤奶时间，一台机器只能挤一头奶牛，至少要多少太机器，输出奶牛使用机器的编号。思路：按照每一头奶牛的开始时间，用优先队列，将奶牛入队，如果前者开始时间大于后者，则需要添加机器。注释理解

#include
#include
#include
#include
using namespace std;
int flag[50005];
struct node
{
 int cow;
 int tim;
 int num;//记录第几个奶牛
 friend bool operator <(node a,node b)
 {
  if(a.tim==b.tim)//时间相同时，结束时间从小到大排，保证优先只用一个机器
  {
   return a.cow  }
  return a.tim>b.tim;
 }
}m[50005];
bool cmp(node x,node y)//排序
{
 if(x.cow!=y.cow)
 return x.cow else
 return x.tim}
priority_queuetemp;
int main()
{
 int n,i,sum;
 while(cin>>n)
 {
  for(i=0;i  {
   cin>>m[i].cow>>m[i].tim;
   m[i].num=i;
  }
  
  sort(m,m+n,cmp);
  sum=1;
  temp.push(m[0]);
  flag[m[0].num]=1;
  for(i=1;i  {
   if(!temp.empty()&&temp.top().tim   {
    flag[m[i].num]=flag[temp.top().num];
    temp.pop();
   }
   else
   {
    sum+=1;//第几个机器
    flag[m[i].num]=sum;
   }
   temp.push(m[i]);
  }
  cout<  for(i=0;i  {
   cout<  }
  while(!temp.empty())
  {
   temp.pop();
  }
 }
 return 0;
}