zoj 3211 Dream City

Dream City

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Time Limit: 1 Second      Memory Limit: 32768 KB

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JAVAMAN is visiting Dream City and he sees a yard of gold coin trees. There aren trees in the yard. Let's call them tree 1, tree 2 ...and tree n. At the first day, each treei has a_i coins on it (i=1, 2, 3...n). Surprisingly, each treei can grow b_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 mostm 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 consecutivem 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 integersn and m (0 < m <= n <= 250) separated by a space. The second line of each test case containsn 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 containsn positive integers separated by a space, indicatingb_i. (0 < b_i <= 100, i=1, 2, 3...n)

Output

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

Sample Input

22 110 101 12 28 102 3

Sample Output

1021

Hints:
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.

思路：按增长速度从小到大排序，然后直接01背包即可

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<cstdlib>#include<stack>#include <queue>#include <set>using namespace std;const int inf = 2147483647;const double PI = acos(-1.0);const int mod = 1000000007;struct node{int a, b, id;}x[300];bool cmp(node a, node b){//if (a.a + a.b == b.a + b.b)return a.b < b.b;//return a.a + a.b < b.a + b.b;}int dp[300];int main(){int t, i, j, n, m;scanf("%d", &t);while (t--){scanf("%d%d", &n, &m);for (i = 1; i <= n; ++i){scanf("%d", &x[i].a);}for (i = 1; i <= n; ++i)scanf("%d", &x[i].b);sort(x + 1, x + 1 + n, cmp);memset(dp, 0, sizeof(dp));for (i = 1; i <= n; ++i){for (j = m; j >= 1; --j){dp[j] = max(dp[j], dp[j - 1] + x[i].a + x[i].b * (j - 1));}}printf("%d\n", dp[m]);}}/*35 51 2 3 4 5 10 5 2 7 1*/