Dream City ZOJ

JAVAMAN is visiting Dream City and he sees a yard of gold coin trees. There are ntrees in the yard. Let's call them tree 1, tree 2 ...and tree n. At the first day, each tree i has a_i coins on it (i=1, 2, 3...n). Surprisingly, each tree i can growb_i new coins each day if it is not cut down. From the first day, JAVAMAN can choose to cut down one tree each day to get all the coins on it. Since he can stay in the Dream City for at most m days, he can cut down at most m trees in all and if he decides not to cut one day, he cannot cut any trees later. (In other words, he can only cut down trees for consecutive m or less days from the first day!)

Given n, m, a_i and b_i (i=1, 2, 3...n), calculate the maximum number of gold coins JAVAMAN can get.

Input

There are multiple test cases. The first line of input contains an integer T (T <= 200) indicates the number of test cases. Then T test cases follow.

Each test case contains 3 lines: The first line of each test case contains 2 positive integers n and m (0 < m <= n <= 250) separated by a space. The second line of each test case contains n positive integers separated by a space, indicating a_i. (0 < a_i <= 100, i=1, 2, 3...n) The third line of each test case also contains npositive integers separated by a space, indicating b_i. (0 < b_i <= 100, i=1, 2, 3...n)

<b< dd="">

Output

For each test case, output the result in a single line.

<b< dd="">

Sample Input

22 110 101 12 28 102 3

<b< dd="">

Sample Output

1021

<b< dd="">

Hint

s:
Test case 1: JAVAMAN just cut tree 1 to get 10 gold coins at the first day.
Test case 2: JAVAMAN cut tree 1 at the first day and tree 2 at the second day to get 8 + 10 + 3 = 21 gold coins in all.

一道比較明显的动态规划，需要注意的是树砍掉之后就不长了，还有需要提前根据树的增长率进行排序，因为要先砍增长率低的树，至于为什么，可以自己想一下。

然后就是状态转移方程，dp[i][j]表示前i棵树在前j天得到的最大值，

dp[i][j] = max( dp[i-1][j] , dp[i-1][j-1] + chushi[i] + zengzhangzhi[i]*(j-1) );

上代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<string>#include<set>using namespace std;struct node{int x;int y;int tmp;};node op[252];int dp[252][252];bool cmp(node a,node b){return a.y < b.y;}int main(){   int t;   scanf("%d", &t);   while(t--)   {     int n,m;     scanf("%d %d",&n,&m);      for(int i=1; i<=n; i++)     {      scanf("%d", &op[i].x); //op[i].tmp = op[i].x;  }  for(int i=1; i<=n; i++)     {      scanf("%d", &op[i].y);   }  memset(dp, 0, sizeof(dp));  sort(op+1, op+1+n,cmp);  for(int i=1; i<=n; i++)  {   for(int j=1; j<=i; j++)   {    dp[i][j] = max(dp[i-1][j], dp[i-1][j-1]+op[i].x+op[i].y*(j-1)); } }  printf("%d\n", dp[n][m]);   }  return 0;} 

水波.