「BZOJ1620」[Usaco2008 Nov] Time Management 时间管理
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
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
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.
题解
按结束时间排序，从后往前扫一遍
注意无解输出-1
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 |
#include<cstdio> #include<algorithm> #include<iostream> #define inf 0x7fffffff using namespace std; int n,now=inf; struct data{int x,y;}a[1001]; bool cmp(data a,data b){return a.y<b.y;} int main() { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y); sort(a+1,a+n+1,cmp); for(int i=n;i>0;i--) now=min(now,a[i].y)-a[i].x; if(now<0)printf("-1"); else printf("%d",now); return 0; } |
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