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個工作,每個工作其所需時間,及完成的Deadline,問要完成所有工作,最遲要什麽時候開始.

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

Sample Output

HINT

Source

Silver

二分最小開始時間

 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){;}

BZOJ——1620: [Usaco2008 Nov]Time Management 時間管理