poj 2376 Cleaning Shifts 最小区间覆盖

Cleaning Shifts

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40751		Accepted: 9871

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 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：有n头牛要值班，每头牛有一个值班区间，要安排牛在[1 , k]区间值班，最少需要几头牛

 

题解：左端点不相等，按从小到大排序，相等按右端点从大到小排序。for循环遍历，在左端点>=now+1的区间中选右端点最大的区间即可

 

#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<vector>
#include<stack>
#include<math.h>
#define mod 998244353
#define ll long long
#define MAX 0x3f3f3f3f
using namespace std;
int vis[];
struct node
{
    int l;
    int r;
}p[];

bool cmp(node a,node b)
{
    if(a.l!=b.l)
        return a.l<b.l;
    else
        return a.r>b.r;
}
int main()
{
    int n,t;
    while(~scanf("%d%d",&n,&t))
    {
        int flag=,cnt=,pos=;
        for(int i=;i<n;i++)
            scanf("%d%d",&p[i].l,&p[i].r);
        sort(p,p+n,cmp);
        if(p[].l!=)
            flag=;
        int now=p[].r;
        for(int i=;i<n&&flag==;)
        {
            if(p[i].l>now+)
            {
                flag=;
                break;
            }
            else
            {
                int temp=now;
                while(i<n&&p[i].l<=now+)//在左端点<=now+1的区间中选右端点最大的区间
                {
                    if(p[i].r>temp)//更新最大右端点
                    {
                        temp=p[i].r;
                        pos=i;
                    }
                    i++;
                }
                if(temp==now)//更新之前和更新之后一样，那还不如不更新
                    continue;
                now=temp;//更新
                cnt++;
            }
        }
        if(p[pos].r!=t)
            flag=;
        if(flag==)
            printf("-1\n");
        else
            printf("%d\n",cnt );
    }
    return ;
}