POJ

Cleaning Shifts

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 10
1 7
3 6
6 10

Sample Output

2

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed. 

INPUT DETAILS: 

There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10. 

OUTPUT DETAILS: 

By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.

 

 

题意：农夫John和他的cows系列。这次cows们被John命令cleaning shifts，给你总区间时间，和每头cow负责clean的子区间时段，问怎样安排可以使最少的cow打扫于整个时间区间。

思路：典型贪心。先对所有cow开始clean的时间升序排序，然后再依次遍历每头cow，保证开始时间在上一头cow时段内，使结束时间最长。（注意：不重叠但相邻也成立）一旦超出上一头cow的结束时段，更新cow，并继续向下遍历。

 

 

#include<stdio.h>
#include<algorithm>
using namespace std;

struct Node{
    int l,r;
}node[25005];
bool cmp(Node a,Node b)
{
    if(a.l==b.l) return a.r>b.r;
    return a.l<b.l;
}
int main()
{
    int n,t,c,f,min,max,i;
    scanf("%d%d",&n,&t);
    for(i=1;i<=n;i++){
        scanf("%d%d",&node[i].l,&node[i].r);
    }
    sort(node+1,node+n+1,cmp);
    c=1;f=0;
    min=node[1].r;
    max=node[1].r; 
    for(i=2;i<=n;i++){
        if(node[i].l<=min+1){    //坑点。。相邻即相连
            if(node[i].r>max){
                f=1;
                max=node[i].r;
                if(max>=t) break;
            }
        }
        else{
            min=max;
            c++;
            f=0;
            if(node[i].l<=min+1){   //同上
                if(node[i].r>max){
                    f=1;
                    max=node[i].r;
                    if(max>=t) break;
                }
            }
            else break;
        }
    }
    if(node[1].l<=1&&max>=t) printf("%d
",c+f);
    else printf("-1
");
    return 0;
}