hdojGirl Love Value 2670 （01背包）

Girl Love Value

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 787    Accepted Submission(s): 442

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

47104
/*题意：一个女神要从n个男生中选m个男生，求最大的爱慕值。 给你两个数n,m，分别代表有n个男生，女神要选m个。接下来一行输入n个数表示每个男生对这个女神的初始爱慕值li，再接着一行输入n个数表示每个男生对这个女神的失望值bi(如果今天女神不选他，他对这个女神的爱慕值就会减少bi。。。)思路：知道题意后解题就不是很难了。就是一个单纯的01背包 */
#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#define ll long long#define N 1010using namespace std;struct zz{int li;int bi;}q[N];int dp[N];int cmp(zz a,zz b){return a.bi>b.bi;}int main(){int n,m,i,j;while(scanf("%d%d",&n,&m)!=EOF){for(i=0;i<n;i++)scanf("%d",&q[i].li);for(i=0;i<n;i++)scanf("%d",&q[i].bi);sort(q,q+n,cmp);memset(dp,0,sizeof(dp));for(i=0;i<n;i++){for(j=m;j>=1;j--){dp[j]=max(dp[j],dp[j-1]+q[i].li-(j-1)*q[i].bi);}}printf("%d\n",dp[m]);}return 0;}