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

1620: [Usaco2008 Nov]Time Management 时间管理

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 636  Solved: 387
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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.

思路:贪心。一眼题。5分钟A掉,爽。 按照s[i]排序,然后完成前i个任务的最晚开始时间为 s[i]-sum ,sum为t[1]+…+t[i]..
如果ans<0,说明无法完成,赋值为-1.

作者: CrazyKK

ex-ACMer@hust,researcher@sensetime

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