HDU - 4221 Greedy? 贪心

iSea is going to be CRAZY! Recently, he was assigned a lot of works to do, so many that you can't imagine. Each task costs Ci time as least, and the worst news is, he must do this work no later than time Di!
OMG, how could it be conceivable! After simple estimation, he discovers a fact that if a work is finished after Di, says Ti, he will get a penalty Ti - Di. Though it may be impossible for him to finish every task before its deadline, he wants the maximum penalty of all the tasks to be as small as possible. He can finish those tasks at any order, and once a task begins, it can't be interrupted. All tasks should begin at integral times, and time begins from 0.

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case includes an integer N. Then N lines following, each line contains two integers Ci and Di.

Technical Specification
1. 1 <= T <= 100
2. 1 <= N <= 100 000
3. 1 <= Ci, Di <= 1 000 000 000

Output

For each test case, output the case number first, then the smallest maximum penalty.

Sample Input

2
2
3 4
2 2
4
3 6
2 7
4 5
3 9

Sample Output

Case 1: 1
Case 2: 3

题意：给出每个任务至少要操作的时间和最晚截止时间，若在t时间完成，t>d,则惩罚为t-d，求所有的惩罚的最大值最小是多少

题解：最后完成所有任务的时间是确定的，要让每个任务的惩罚尽可能小，当然要先做截止时间靠前的，因为若放在后面，完成的时间越靠后，惩罚越大，这样最大值会更大，按截止时间排序，记录下当前时间即可

#include
#include
#include
#include
using namespace std;
const int N=110000;
typedef long long ll;
struct node{
	ll x;
	ll end;
}a[N];
bool cmp(node xx,node yy)
{
	return xx.enda[i].end)
			{
				ans=max(ans,tm-a[i].end);
			}
		}
		printf("Case %d: %lld\n",nn++,ans);
	}
	return 0;
}