hdu 1789 Doing Homework again 贪心

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20510    Accepted Submission(s): 11813

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

题意：给你完成每一个作业的截止日期，和每一个超过截至日期扣的分数，规定每天只能完成一项作业，要求完成所有作业使扣的总分数最小

 

题解：按扣分从大到小排序，若扣分相同，截止日期小的排在前面；按排序后的顺序处理作业，从该作业的截至日期开始往前遍历，若还有某天未做作业（vis[i]==0）,则该作业不计罚分

 

#include<iostream>
#include<algorithm>
#include<string.h>
#define ll long long 
using namespace std;
struct node
{
    ll t;
    ll s;
}p[1005];
bool cmp(node a,node b)
{
    if(a.s!=b.s)
        return a.s>b.s;
    else
        return a.t<b.t;
}
int vis[1005];
int main()
{
    int t,n;
    cin>>t;
    while(t--)
    {
        
        cin>>n;
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
            cin>>p[i].t;
        
        for(int i=1;i<=n;i++)
            cin>>p[i].s;
        sort(p+1,p+1+n,cmp);
        ll ans=0;
        for(int i=1;i<=n;i++)
        {
            int flag=0;
            for(int j=p[i].t;j>=1;j--)
            {
                if(vis[j]==0)
                {
                    vis[j]=1;
                    flag=1;
                    break;
                }
            }
            if(flag==0)
                ans=ans+p[i].s;
        }
        cout<<ans<<endl;

    }
    return 0;

}