[BZOJ] 1620: [Usaco2008 Nov]Time Management 时间管理

1620: [Usaco2008 Nov]Time Management 时间管理

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 850  Solved: 539
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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

4
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

2

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

Silver

 

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 }
心情极差qwq
posted @ 2017-09-19 20:14  Leviaton  阅读(158)  评论(0编辑  收藏  举报