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.
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 only one integer about the maximum total Love value Kiki can get by choose K boys.
3 3
10 20 30
4 5 6
4 3
20 30 40 50
2 7 6 5
47
104
题解： 注意在选的这些个人中，肯定是每日损失最大的先选，所以按照损失从大到小排序，然后就是一个简单的0,1排序了。
dp[i][j] = max(dp[i-1][j],dp[i-1][j-1]+a[i]-b[i]*(j-1);
可以压缩成一维的
代码：
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 using namespace std;
5 const int N = 1010;
6 int dp[N];
7 struct Node{
8 int a;
9 int b;
10 bool operator < (const Node& tm) const{
11 return b>tm.b;
12 }
13 }node[N];
14 int main()
15 {
16 int n,k;
17 while(~scanf("%d%d",&n,&k))
18 {
19 for(int i = 0; i < n; i++)
20 scanf("%d",&node[i].a);
21 for(int i = 0; i < n; i++)
22 scanf("%d",&node[i].b);
23 memset(dp,0,sizeof(dp));
24 sort(node,node+n);
25 for(int i = 0; i < n; i++)
26 {
27 for(int j = k; j > 0; j--)
28 {
29 dp[j] = max(dp[j],dp[j-1]+node[i].a-node[i].b*(j-1));
30 }
31 }
32 printf("%d\n",dp[k]);
33 }
34 return 0;
35 }
