Cleaning Shifts

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 2

Problem Description

Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T.

Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval.

Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.

 

 

Input

* Line 1: Two space-separated integers: N and T <br> <br>* Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.

 

 

Output

* Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.

 

 

Sample Input

3 10 1 7 3 6 6 10

 

 

Sample Output

2

 

每头牛有一定的工作时间段，问最少用多少头牛才能保证在所有时间内都有牛在工作

用贪心算法，每次选取工作结束时间最晚的牛。

记住：

2 10

1 5

1 6

也是可以的，不一定要1 5和5 10

 

#include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             using namespace std;struct node{    int s, e;}a[1000005];bool cmp(node a, node b)//贪心{    if (a.s == b.s)//开始时间短的在前        return a.e
             
              > n >> t; for (i = 1; i <= n; i++) { cin >> a[i].s >> a[i].e; } a[n + 1].s = 999999999;//最后一个开始时间设为无穷大 sort(a + 1, a + 1 + n, cmp); int mm = 0; int ans = 0; bool f = 0; int ss =0; for (i = 1; i <= n; i++) { if (a[i].s <=ss+1 )//是ss+1 { if (a[i].e > mm) { mm = a[i].e; f = 1; } if (a[i + 1].s > ss+1 && f == 1) { ss = mm; ans++; f = 0; } } } if (ss>=t) cout << ans << endl; else cout << "-1" << endl; return 0;}