hdoj 1789 Doing Homework again

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7702    Accepted Submission(s): 4562

Problem Description

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

 

Input

The input contains several test cases. The first line of the input is a single integer T that is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=1000) which indicate the number of homework.. Then 2 lines follow. The first line contains N integers that indicate the deadlines of the subjects, and the next line contains N integers that indicate the reduced scores.

 

Output

For each test case, you should output the smallest total reduced score, one line per test case.

 

Sample Input

   
   
   
   

    
    
    
    3
3
3 3 3
10 5 1
3
1 3 1
6 2 3
7
1 4 6 4 2 4 3
3 2 1 7 6 5 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0
3
5
   
   
   
   

 

贪心：引进数组use ，用use数组来表示第  i  天是否被占用，若占用为  use [ i ]  =  1，否则为 0 ；

先按分值降序排列，每次优先考虑分值大的，从该分值的截止日期递推，若可以找到空闲的一天 j（即use [ j ] = 0） ，则记录use [ j ] = 1并跳出循环进行下一个分值判定，若找不到空闲的一天说明该分值需要抛弃 。

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define max 1000+10
using namespace std;
int use[max];
struct record
{
    int time;
    int score;
}num[max];
bool cmp(record a,record b)
{
    return a.score>b.score;
}
int main()
{
    int t,i,j;
    int work,mincost;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&work);
        memset(use,0,sizeof(use));//初始化 
        for(i=0;i<work;i++)
        scanf("%d",&num[i].time);
        for(i=0;i<work;i++)
        scanf("%d",&num[i].score);
        sort(num,num+work,cmp);
        mincost=0;
        for(i=0;i<work;i++)
        {
            for(j=num[i].time;j>=1;j--)
            {
                if(use[j]==0)
                {
                    use[j]=1;
                    break;
                }
            }
            if(j==0)//找不到空闲的一天 
            mincost+=num[i].score;
        }
        printf("%d\n",mincost);
    }
    return 0;
}