hdu2670

Girl Love Value

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 880    Accepted Submission(s): 510

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

3 310 20 304 5 64 320 30 40 502 7 6 5

 

Sample Output

47
104
AC代码：
#include <ctime>#include <string>#include<cstdio>#include<algorithm>using namespace std;int n,k;struct point{        int a,b;}p[100009];bool cmp(point x,point y){        return x.b>y.b;}int dp[100009][1009];int main(){        while(scanf("%d%d",&n,&k)!=EOF)        {                for(int i=1;i<=n;i++)                        scanf("%d",&p[i].a);                for(int i=1;i<=n;i++)                        scanf("%d",&p[i].b);                int j=0,ans=0;                sort(p+1,p+n+1,cmp);                for(int i=0;i<=n;i++)                        dp[i][0]=0;                for(int i=1;i<=n;i++)                {                        for(int j=1;j<=i&&j<=k;j++)                                dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]+(p[i].a-(j-1)*p[i].b));                }                printf("%d\n",dp[n][k]);        }        return 0;}