USACO2008 Time Management /// 贪心 oj24386

题目大意：

有N个工作被编号为1..N (1 ≤ N ≤ 1,000) 

完成第i个工作需要T_i (1 ≤ T_i ≤ 1,000)的时间

第i个工作需在S_i (1 ≤ S_i ≤ 1,000,000)前结束

若能按时完成则输出 最晚开始工作的时间  若不能则输出 -1

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

Sample Output

2

Hint

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.

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.

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
struct job
{
    int s,e;
}a[1005];
bool cmp(struct job a,struct job b)
{
    return a.e<b.e;
}
int main()
{
    int n;
    scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d%d",&a[i].s,&a[i].e);
        sort(a+1,a+1+n,cmp); /// 按最晚结束时间排序
        int ans=INF;
        for(int i=n;i>=1;i--) /// 从最晚结束的事件开始遍历
            ans=min(ans,a[i].e)-a[i].s;
        /* 最晚开始时间与前一件事的最晚结束时间取更早的一个
                 最终推出第一件事的最晚开始时间
         若无法按时完成 则时间会被推到0之前 也就是ans<0 */
        if(ans<0) printf("-1
");
        else printf("%d
",ans);
    return 0;
}