poj 2376 Cleaning Shifts 最小区间覆盖
Cleaning Shifts
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 40751 | Accepted: 9871 |
Description
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
* 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
Sample Input
3 10
1 7
3 6
6 10
Sample Output
2
Hint
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.
#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 ;
}
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