A - Doing Homework again

A - Doing Homework again

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

333 3 310 5 131 3 16 2 371 4 6 4 2 4 33 2 1 7 6 5 4 

 

Sample Output

035 

代码：

#include<stdio.h>
#include<algorithm>
#include <cstring>
using namespace std;
struct work
{
    int data;
    int score;
}a[1005];
int comp(struct work  a,struct work b)
{
    if(a.score!=b.score)//首先按照分数不同高->低排序
        return a.score>b.score;
    if(a.score==b.score)//分数相同时，日期小的先排序，因为日期小的可以提前做完再考虑日期大的，以便日期大的往前推 的时候不用再考虑日期小的有没有做完
    return a.data<b.data;
}
int main()
{
    int t,Find[1005];
    scanf("%d",&t);
    while(t--)
    {
        int i,n;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i].data);
        }
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i].score);
        }
        int cnt=0;
        sort(a,a+n,comp);
        memset(Find,0,sizeof(Find));
        for(i=0;i<n;i++)
        {
            int day=a[i].data;
            while(day)
            {
                if(!Find[day])// 从截止时间开始往前推,如果有一天没用过,这一天就做这一门课,这门课不扣分
                {
                    Find[day]=1;
                    break;
                }
                day--;
            }
            if(day==0)
                cnt+=a[i].score;
        }
        printf("%d\n",cnt);
    }
    return 0;
}