【BZOJ1630】【Usaco2008 Nov】【Time Management 时间管理】【贪心】

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.

题解：这个题意真是醉了。。。

看懂题之后从最后结束的时间往前推即可。如果最后的得到的开始时间小于0就输出-1.

代码：

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;struct use{int st,en;}a[1000010];int n,ans(99999999);bool cmp(use a,use b) {return a.en>b.en;}int main(){    scanf("%d",&n);    for (int i=1;i<=n;i++){scanf("%d%d",&a[i].st,&a[i].en);}    sort(a+1,a+n+1,cmp);    for (int i=1;i<=n;i++)ans=min(ans,a[i].en)-a[i].st;    if (ans<0) ans=-1;    cout<<ans<<endl; }