HDU 1789 Doing Homework again（排序，DP）

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2969    Accepted Submission(s): 1707

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

 

Author

lcy

 

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

 

Recommend

lcy

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int MAXN=1010;

struct Node
{
    int d,s;
}node[MAXN];

bool used[10000];

bool cmp(Node a,Node b)
{
    if(a.s==b.s)
    {
        return a.d<b.d;
    }    
    return a.s>b.s;
}      //按减分从大到小排，按日期从小到大排  

int main()
{
    int T;
    int n;
    int j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%d",&node[i].d);
        for(int i=0;i<n;i++) scanf("%d",&node[i].s);
        sort(node,node+n,cmp);
        memset(used,false,sizeof(used));//装每天所学的课程
        int ans=0;
        for(int i=0;i<n;i++)
        {
            for(j=node[i].d;j>0;j--)//从截止日期开始往前数
            {
                if(!used[j])
                {
                    used[j]=true;
                    break;
                }    
            }    
            if(j==0)//当一个位置都没有，就减分
              ans+=node[i].s;
        }    
        printf("%d
",ans);
    }    
    return 0;
}

都说这个是水题，但是我想了很久。。。。。

继续努力吧！