HDU 4221 Greedy?(贪心)

Greedy?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 908    Accepted Submission(s): 284

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

 

Author

iSea@WHU

 

Source

首届华中区程序设计邀请赛暨第十届武汉大学程序设计大赛

 

Recommend

lcy

 

 

 

贪心。

就是按照结束时间从小到大排序。之后按照这个顺序去求结果就可以了。

至于为什么这样贪心是正确的，我也还没有想明白。

#include<stdio.h>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int MAXN=100010;
struct Node
{
    int c,d;
}node[MAXN];
bool cmp(Node a,Node b)
{
    return a.d<b.d;
}
int main()
{
    int T;
    int iCase=0;
    scanf("%d",&T);
    int n;
    while(T--)
    {
        iCase++;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
          scanf("%d%d",&node[i].c,&node[i].d);
        sort(node,node+n,cmp);
        long long sum=0;
        long long ans=0;
        for(int i=0;i<n;i++)
        {
            sum+=node[i].c;
            if(sum-node[i].d>ans)ans=sum-node[i].d;
        }
        printf("Case %d: %I64d\n",iCase,ans);
    }
    return 0;
}