Cleaning Shifts - POJ 2376 贪心

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Cleaning Shifts
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12574 Accepted: 3257


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.


* 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.


* 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




#include<iostream>#include<cstdio>#include<cstdlib>#include<string>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<stack>#define INF 0x777777#define MAXN 25010using namespace std;struct node{int s,e;}cow[MAXN];bool cmp(node a,node b){return a.s<b.s;}int main(){int n,T;scanf("%d%d",&n,&T);for(int i=1;i<=n;i++)scanf("%d%d",&cow[i].s,&cow[i].e);sort(cow+1,cow+1+n,cmp);if(cow[1].s>1) {printf("-1\n");return 0;}int start=0,end=0,cnt=0; //注意start和end的初始条件for(int i=1;i<=n;i++){if(cow[i].s<=start)//如果小于等于start只需更新end end=max(end,cow[i].e);else{//大于start则增加计数,此时保证取的是end最远的线段 cnt++;start=end+1;//只需从end的下一个点开始 if(cow[i].s<=start)end=max(end,cow[i].e);//不要忘记更新end else{printf("-1\n");return 0;}}if(end>=T) break;}if(end>=T)printf("%d\n",cnt);elseprintf("-1\n");return 0;}

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