动态规划： HDU 1789Doing Homework again

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

 

可以将作业按扣分高低排序，每次找到扣分最高的，将它安排进终止日期那一天，如果那天有事，就在往前推，，如果最后安排不下，就扣分。这样循环n次就可以了。

sort排序。。。。

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
#define N 1100
using namespace std;

bool hh[N];
typedef struct nod{
    int x,y;
}s[N];

bool cmp(nod a,nod b)
{
    if(a.y!=b.y)
        return a.y>b.y;
    else
        return b.x>a.x;
}
int main()
{
    nod s[N];
    int T,n,i,j,sum;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&s[i].x);
        }
        for(i=1;i<=n;i++)
            scanf("%d",&s[i].y);
        sort(s+1,s+n+1,cmp);
        sum=0;
        memset(hh,false,sizeof(hh));
        for(i=1;i<=n;i++)
        {
            for(j=s[i].x;j>0;j--)
            {
                if(hh[j]==false)
                {
                    hh[j]=true;
                    break;
                }
            }
            if(j==0)
            {
                sum=sum+s[i].y;
            }
        }
        printf("%d
",sum);
    }
    return 0;
}