hdu4221 Greedy? _贪心

对n项任务按d由小到大排序，然后计算出其中的最大处罚即为所求

 

 

Greedy?

 

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2162    Accepted Submission(s): 733

 

 

Problem Description

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：完成一个任务的需要的时间为Ci，这个任务完成的最后期限为Di，如果完成时间Ti超过Di，
就得到Ti-Di的处罚。如何使完成所有任务后的最大处罚尽可能小 ，并输出最小的最大处罚。 
*/ 

/*
分析：假设只有两个任务1、2，且d1c1-d1,c1+c2-d1>c1+c2-d2,所以应先完成截止时间早的任务
	  
	考虑有3个任务，假设d1
#include
#include 
using namespace std;

struct time
{
	int c,d;
}task[100010];

bool cmp(time a,time b)
{
	return a.d>t;
	for(int k=1;k<=t;k++)
	{
		int n;
		cin>>n;
		for(int i=0;imin_max)
			  min_max=sum_time-task[i].d;
		}
		printf("Case %d: %lld\n",k,min_max);
	}
	return 0;
}