HDU 2670 Girl Love Value



Girl Love Value

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 743    Accepted Submission(s): 412

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

 

Author

yifenfei

 

Source

HDU女生专场公开赛——谁说女子不如男

题意：有n个男生追一个女生，每个男生有一个相应的爱慕值Vx（i），给定一个天数m，每天这个女生只能找一个女生，并且得到这个男生的爱慕值。每过一天，每个男生的爱慕值就会减少相应的 Vy（i）；问女生如何选择，m天之后，才能得到最大的爱慕值。

咋看一眼是贪心的题目，每次都先选爱慕值最大的，和这个女生约会。。可是后来发现问题，要是天数很多的话，有些男生的爱慕值减少的很快，这样贪心就不好处理了，因为不清楚m天后的爱慕值，也不知道是先选减得多的还是爱慕值大的。

其实是01背包的变形。

讨论选不选第n-1个人。

假如选，爱慕值为dp[j]。假如不选，爱慕值为dp[j-1]+这个男生的爱慕值-（每天减少的爱慕值*天数）。

上代码

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include <math.h>using namespace std;struct node{    int value,ans;};bool cmp(node a,node b){    return a.ans>b.ans;}/*---------------------------------------------*/   //下面注释部分为贪心的错误解法。/*---------------------------------------------*/int main(){    node f[1005];    int dp[1005];    int n,day,i,j,love,Max;    while(cin>>n>>day)    {        memset(dp,0,sizeof(dp));        // love=0;        for(i=0; i<n; i++)            cin>>f[i].value;        for(i=0; i<n; i++)            cin>>f[i].ans;        sort(f,f+n,cmp);        /*        while(day--)        {                love+=f[0].value;                f[0].value=-100005;                break;            }            for(i=0; i<n; i++)            {                if(f[i].value!=-100005)                f[i].value-=f[i].ans;            }            sort(f,f+n,cmp);        }        */        for(i=0; i<n; i++)            for(j=day; j>0; j--)            {                dp[j]=max(dp[j],dp[j-1]+f[i].value-f[i].ans*(j-1));            }        // cout<<love<<endl;        cout<<dp[day]<<endl;    }    return 0;}