[BZOJ] 1620: [Usaco2008 Nov]Time Management 时间管理
Posted Lev今天学习了吗
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了[BZOJ] 1620: [Usaco2008 Nov]Time Management 时间管理相关的知识,希望对你有一定的参考价值。
1620: [Usaco2008 Nov]Time Management 时间管理
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 850 Solved: 539
[Submit][Status][Discuss]
Description
Ever the maturing businessman, Farmer John realizes that he must manage his time effectively. He has N jobs conveniently numbered 1..N (1 <= N <= 1,000) to accomplish (like milking the cows, cleaning the barn, mending the fences, and so on). To manage his time effectively, he has created a list of the jobs that must be finished. Job i requires a certain amount of time T_i (1 <= T_i <= 1,000) to complete and furthermore must be finished by time S_i (1 <= S_i <= 1,000,000). Farmer John starts his day at time t=0 and can only work on one job at a time until it is finished. Even a maturing businessman likes to sleep late; help Farmer John determine the latest he can start working and still finish all the jobs on time.
N个工作,每个工作其所需时间,及完成的Deadline,问要完成所有工作,最迟要什么时候开始.
Input
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains two space-separated integers: T_i and S_i
Output
* Line 1: The latest time Farmer John can start working or -1 if Farmer John cannot finish all the jobs on time.
Sample Input
3 5
8 14
5 20
1 16
INPUT DETAILS:
Farmer John has 4 jobs to do, which take 3, 8, 5, and 1 units of
time, respectively, and must be completed by time 5, 14, 20, and
16, respectively.
Sample Output
OUTPUT DETAILS:
Farmer John must start the first job at time 2. Then he can do
the second, fourth, and third jobs in that order to finish on time.
HINT
Source
Analysis
一波排序,把死线从后往前排
只有最后一条死线可以压着线做
然而由于农夫超~~~~专注的,同一时间只能做一件事
因此从后往前遍历,往时间轴上安排事情
防止前面的事情影响后面压线完成任务
=w=
细节自行思考
Code
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #define maxn 10000000 5 using namespace std; 6 7 struct job{ 8 int t,deadline; 9 }list[maxn/1000]; 10 11 int n; 12 13 int cnt[maxn]; 14 15 bool cmp(const job &a,const job &b){ 16 return a.deadline < b.deadline; 17 } 18 19 int main(){ 20 scanf("%d",&n); 21 22 for(int i = 1;i <= n;i++) 23 scanf("%d%d",&list[i].t,&list[i].deadline); 24 25 sort(list+1,list+1+n,cmp); 26 27 for(int i = n;i >= 1;i--){ 28 int j = list[i].deadline; 29 while(cnt[j] && j) j--; 30 list[i].deadline = j; 31 int tmp = list[i].t; 32 while(tmp && j) cnt[j--] = 1,tmp--; 33 } 34 35 int ans = list[1].deadline-list[1].t; 36 if(ans < 0) printf("-1"); 37 else printf("%d",ans); 38 // cout << ans; 39 40 // for(int i = 1;i <= n;i++){ 41 // printf("%d %d\n",list[i].t,list[i].deadline); 42 // } 43 44 45 46 return 0; 47 }
以上是关于[BZOJ] 1620: [Usaco2008 Nov]Time Management 时间管理的主要内容,如果未能解决你的问题,请参考以下文章
[BZOJ1620][Usaco2008 Nov]Time Management 时间管理
[BZOJ] 1620: [Usaco2008 Nov]Time Management 时间管理
BZOJ_1620_[Usaco2008_Nov]_Time_Management_时间管理_(二分+贪心)