hdoj2670Girl Love Value【01背包】

Girl Love Value Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 769    Accepted Submission(s): 431 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   Author yifenfei

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1010;
int MIM(int a,int b){
return a<b?a:b;
}
int MAX(int a,int b){
return a>b?a:b;
}
struct Node{
int l,b;
}A[maxn];
bool cmp(Node a,Node b){
return a.b>b.b;
}
int dp[maxn];
int main()
{
int i,j,k,n;
while(scanf("%d%d",&n,&k)!=EOF){
for(i=0;i<n;++i){
scanf("%d",&A[i].l);
}
for(i=0;i<n;++i){
scanf("%d",&A[i].b);
}
sort(A,A+n,cmp);//保证优先选择当前最大的
memset(dp,0,sizeof(dp));
for(i=0;i<n;++i){
for(j=k;j>=1;--j){
dp[j]=MAX(dp[j],dp[j-1]+A[i].l-A[i].b*(j-1));
}
}
printf("%d\n",dp[k]);
}
return 0;
}