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

Posted 2020-10-14

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 920  Solved: 569
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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

 

二分最小开始时间

 1 #include <algorithm>
 2 #include <cstdio>
 3 
 4 inline void read(int &x)
 5 {
 6     x=0; register char ch=getchar();
 7     for(; ch>‘9‘||ch<‘0‘; ) ch=getchar();
 8     for(; ch>=‘0‘&&ch<=‘9‘; ch=getchar()) x=x*10+ch-‘0‘;
 9 }
10 const int N(100005);
11 struct Work {
12     int t,s;
13     bool operator < (const Work&x)const
14     {
15         if(s==x.s) return t<x.t;
16         return s<x.s;
17     }
18 }job[N];
19 
20 int ans=-1,L,R,Mid,n;
21 inline bool check(int x)
22 {
23     for(int i=1; i<=n; ++i)
24     {
25         if(x+job[i].t>job[i].s) return 0;
26         x+=job[i].t;
27     }
28     return 1;
29 }
30 
31 int Presist()
32 {
33 //    freopen("manage.in","r",stdin);
34 //    freopen("manage.out","w",stdout);
35     
36     read(n);
37     for(int t,i=1; i<=n; ++i)
38         read(job[i].t),read(job[i].s);
39     std::sort(job+1,job+n+1);
40     for(R=job[1].s-job[1].t+1; L<=R; )
41     {
42         Mid=L+R>>1;
43         if(check(Mid))
44         {
45             ans=Mid;
46             L=Mid+1;
47         }
48         else R=Mid-1;
49     }
50     printf("%d\n",ans);
51     return 0;
52 }
53 
54 int Aptal=Presist();
55 int main(int argc,char**argv){;}