[DP problem] girl love value (DP problem) hdu2670

Problem descriptionLove in college is a happy thing but always have so far 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 beauti Ful girl !!!!. Now you must be sure how wonderful things it is if you are 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. InputThe 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)

OutputOutput 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

47104

Resolution: Sort Bi values in ascending order and then perform DP. DP [I] [J] indicates the maximum value of J for the first I.

DP [I] [J] = max (DP [I-1] [J], DP [I-1] [J-1] + J select the value of I)

#include<stdio.h>#include<stdlib.h>#include<string.h>typedef struct da  {   int li;   int bi;}da;da dd[1001];int dp[1001][1001];int cmp(const void *a,const void *b){   da * c=(da *)a;   da * d=(da *)b;   return d->bi-c->bi;}int main(){int i,j,n,k;  while (scanf("%d%d",&n,&k)!=EOF)  {  for(i=1;i<=n;i++)  scanf("%d",&dd[i].li);  for(i=1;i<=n;i++)  scanf("%d",&dd[i].bi);  qsort(dd+1,n,sizeof(dd[0]),cmp);  memset(dp,0,sizeof(dp));  for(i=1;i<=n;i++)  for(j=1;j<=i;j++)  {  if(i-1>=j) dp[i][j]=dp[i-1][j];  int temp=dp[i-1][j-1]-dd[i].bi*(j-1)+dd[i].li;  if(i-1>=j-1&&temp>dp[i][j])  dp[i][j]=temp;  }  printf("%d/n",dp[n][k]);  }}/*3 310 20 304 5 64 320 30 40 502 7 6 5*/