HDU 1789 Doing Homework again(ソート,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省大会合宿チーム練習試合(10)DOOMIIIに感謝します
 
 
Recommend
lcy
 
 
 
まず、減点の大きい順に並べ替え、点数が同じなら締め切りの小さい順に並べ替えます.
 
そして順番に、締め切り日から先の方に時間がかかっていない時間を探します.
見つからなかったらペナルティに加算します
 
具体的にプログラムを見る.

#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;

}        



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));

        int ans=0;

        for(int i=0;i<n;i++)

        {

            for(j=node[i].d;j>0;j--)

            {

                if(!used[j])

                {

                    used[j]=true;

                    break;

                }    

            }    

            if(j==0)

              ans+=node[i].s;

        }    

        printf("%d
",ans);

    }    

    return 0;

}