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

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.

* Line 1: A single integer: N

* Lines 2..N+1: Line i+1 contains two space-separated integers: T_i and S_i

* Line 1: The latest time Farmer John can start working or -1 if Farmer John cannot finish all the jobs on time.

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