POJ 2376 Cleaning Shifts（贪心）-白红宇

POJ 2376 Cleaning Shifts（贪心）

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发布时间：2019-06-20

本文共 1628 字，大约阅读时间需要 5 分钟。

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

* 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

3 101 73 66 10

Sample Output

最小代价的区间覆盖。

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          using namespace std;typedef long long LL;const int maxn=25000+100;struct node{ int st,ed;}e[maxn];int n,m;int cmp(node l1,node l2){ if(l1.st==l2.st) return l1.ed>l2.ed; return l1.st
           
            ed+1) { ed=temp; ans++; } if(e[i].st<=ed+1) { if(e[i].ed>temp) temp=e[i].ed; if(e[i].ed==m) { ans++; ed=m; break; } } } if(ed==m) printf("%d\n",ans); else printf("-1\n"); } return 0;}

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