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
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){;}
