hdu_2670Girl Love Value(dp)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2670

Girl Love Value

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 795    Accepted Submission(s): 448

Problem Description

 

Love in college is a happy thing but always have so many pity boys or girls can not find it.
Now a chance is coming for lots of single boys. The Most beautiful and lovely and intelligent girl in HDU,named Kiki want to choose K single boys to travel Jolmo Lungma. You may ask one girls and K boys is not a interesting thing to K boys. But you may not know Kiki have a lot of friends which all are beautiful girl!!!!. Now you must be sure how wonderful things it is if you be choose by Kiki.



Problem is coming, n single boys want to go to travel with Kiki. But Kiki only choose K from them. Kiki every day will choose one single boy, so after K days the choosing will be end. Each boys have a Love value (Li) to Kiki, and also have a other value (Bi), if one boy can not be choose by Kiki his Love value will decrease Bi every day.
Kiki must choose K boys, so she want the total Love value maximum.

 

 

Input

 

The input contains multiple test cases.
First line give the integer n,K (1<=K<=n<=1000)
Second line give n integer Li (Li <= 100000).
Last line give n integer Bi.(Bi<=1000)

 

 

Output

 

Output only one integer about the maximum total Love value Kiki can get by choose K boys.

 

 

Sample Input

 

3 3 10 20 30 4 5 6 4 3 20 30 40 50 2 7 6 5

 

 

Sample Output

 

47 104

题解： 注意在选的这些个人中，肯定是每日损失最大的先选，所以按照损失从大到小排序，然后就是一个简单的0,1排序了。

dp[i][j] = max(dp[i-1][j],dp[i-1][j-1]+a[i]-b[i]*(j-1);

可以压缩成一维的

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1010;
int dp[N];
struct Node{
    int a;
    int b;
    bool operator < (const Node& tm) const{
        return b>tm.b;
    }
}node[N];
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        for(int i = 0; i < n; i++)
            scanf("%d",&node[i].a);
        for(int i = 0; i < n; i++)
            scanf("%d",&node[i].b);
        memset(dp,0,sizeof(dp));
        sort(node,node+n);
        for(int i = 0; i < n; i++)
        {
            for(int j = k; j > 0; j--)
            {
                dp[j] = max(dp[j],dp[j-1]+node[i].a-node[i].b*(j-1));
            }
        }
        printf("%d\n",dp[k]);
    }
    return 0;
}