Stall Reservations

Posted 2020-10-22 如有一味绝境，非历十方生死

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。

 1 struct node{
 2     int a;
 3     int b;
 4     int flag;
 5      friend bool operator <(node node1,node node2)  
 6     {  
 7         //<为从大到小排列，>为从小到大排列  
 8         return node1.b<node2.b;  
 9     }  
10 }n[5];

 1 #include<iostream>
 2 #include<algorithm> 
 3 #include<queue>
 4 using namespace std;
 5 
 6 struct node{
 7     int a; //开始时间 
 8     int b; //结束时间 
 9     int c; //序列号 
10     int flag;    //区间安排序号 
11     friend bool operator <(node node1,node node2) //重载比较操作符函数 
12     {
13         return node1.b > node2.b;
14     }
15 }cows[50005];
16 
17 bool cmp1(node t1,node t2)  //根据区间排序 
18 {
19     if(t1.a != t2.a) return t1.a < t2.a;
20     else return t1.b < t2.b;
21 }
22 
23 bool cmp2(node t1,node t2) //根据序列号排序 
24 {
25      return t1.c < t2.c;
26 }
27 
28 
29 int main()
30 {
31     int n;
32     priority_queue<node>que;
33     while(~scanf("%d",&n))
34     {
35         for(int i = 0; i < n; i++)
36         {
37             scanf("%d %d",&cows[i].a,&cows[i].b);
38             cows[i].c = i;
39         }
40         sort(cows,cows+n,cmp1);  
41         int  ans = 1;
42         cows[0].flag = ans;
43         que.push(cows[0]);
44         
45         for(int i = 1 ; i < n; i++)
46         {
47             if(que.top().b >= cows[i].a)  //此时cows的区间与que内的任意区间有重合 
48             {
49                 ans++;
50                 cows[i].flag = ans;
51             }
52             else
53             {
54                 cows[i].flag = que.top().flag;
55                 que.pop();
56             }
57             que.push(cows[i]);
58         }
59 
60         sort(cows,cows+n,cmp2);  
61         cout<<ans<<endl;
62         for(int i = 0; i < n; i++)
63         {
64             printf("%d\n",cows[i].flag);
65         }
66     }
67     return 0;
68 }