BZOJ——1620: [Usaco2008 Nov]Time Management 时间管理
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
43 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
2OUTPUT 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
二分最小开始时间
#include <algorithm> #include <cstdio> inline void read(int &x) { x=0; register char ch=getchar(); for(; ch>'9'||ch<'0'; ) ch=getchar(); for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0'; } const int N(100005); struct Work { int t,s; bool operator < (const Work&x)const { if(s==x.s) return t<x.t; return s<x.s; } }job ; int ans=-1,L,R,Mid,n; inline bool check(int x) { for(int i=1; i<=n; ++i) { if(x+job[i].t>job[i].s) return 0; x+=job[i].t; } return 1; } int Presist() { // freopen("manage.in","r",stdin); // freopen("manage.out","w",stdout); read(n); for(int t,i=1; i<=n; ++i) read(job[i].t),read(job[i].s); std::sort(job+1,job+n+1); for(R=job[1].s-job[1].t+1; L<=R; ) { Mid=L+R>>1; if(check(Mid)) { ans=Mid; L=Mid+1; } else R=Mid-1; } printf("%d\n",ans); return 0; } int Aptal=Presist(); int main(int argc,char**argv){;}
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