POJ
来源:互联网 发布:sql server2005卸载 编辑:程序博客网 时间:2024/06/05 17:58
题目链接:http://poj.org/problem?id=2376点击打开链接
Cleaning Shifts
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23408 Accepted: 5853
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.
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.
* 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
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.
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.
Source
USACO 2004 December Silver
给你一些区间 求覆盖1~T最少区间个数
贪心思想 以开始时间排序 然后维护当前最远距离maxending 当前终点nowending
每次结束nowending将maxending的值赋给nowding 就能计算最小个数
注意这道题为开区间 端点可以不重合
写法很多 建议用for循环写
while写崩了。。
#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<queue>#include<map>#include<math.h>#include<limits.h>#include<vector>using namespace std;vector <pair<long long int,long long int > >s;int main(){ long long int n,t; scanf("%lld%lld",&t,&n); for(int i=0;i<t;i++) { int mid1,mid2; scanf("%d%d",&mid1,&mid2); if(mid1>mid2) swap(mid1,mid2); s.push_back(make_pair(mid1, mid2)); } sort(s.begin(),s.end()); s.push_back(make_pair(INT_MAX, INT_MAX)); int flag=0; int cnt=0; long long int maxending=0; long long int nowending=0; for(int i=0;i<s.size();i++) { if(s[i].first<=nowending+1) { if(s[i].second>maxending) { maxending=s[i].second; flag=1; } if(s[i+1].first>nowending+1&&flag) { nowending=maxending; cnt++; flag=0; } } } if(nowending<n) cout << "-1"; else cout << cnt;}
阅读全文
0 0
- POJ
- poj
- POJ
- POJ
- poj
- poj
- POJ
- POJ
- poj
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- POJ
- Tornado
- java 注解理解
- react-native <WebView>内部网页跳转并返回上一级
- python在子线程中使用pyHook监控键盘无效,需要加pythoncom的初始化
- Token原理以及应用
- POJ
- Bootstrap概述
- 指针数组,数组指针,二维数组的动态内存分配
- [bzoj1651][Usaco2006 Feb]Stall Reservations 专用牛棚
- 企业如何迎战巨量 DDoS 攻击?轻松两招有备无患
- CTF writeup: rbash 逃脱方法
- Spring、Spring MVC、Struts2、、优缺点整理
- Linux-系统文件
- liunx 定位IO瓶颈方法