博客
关于我
ZOJ3211 Dream City(DP)
阅读量:396 次
发布时间:2019-03-05

本文共 1920 字,大约阅读时间需要 6 分钟。

JAVAMAN is visiting Dream City and he sees a yard of gold coin trees. There are n trees in the yard. Let’s call them tree 1, tree 2 …and tree n. At the first day, each tree i has ai coins on it (i=1, 2, 3…n). Surprisingly, each tree i can grow bi 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, ai and bi (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 ai. (0 < ai <= 100, i=1, 2, 3…n) The third line of each test case also contains n positive integers separated by a space, indicating bi. (0 < bi <= 100, i=1, 2, 3…n)

Output

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

Sample Input

2

2 1
10 10
1 1
2 2
8 10
2 3
Sample Output

10

21
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:把b【i】比较大的尽量放到后面,dp【i】【j】表示前i棵树第j天的最大价值。

#include
using namespace std;typedef long long ll;const int maxn=3e2+1;const int inf=0x3f3f3f3f;int dp[maxn][maxn];pair
p[maxn];bool cmp(const pair
&a,const pair
&b){ return a.second==b.second?a.first

转载地址:http://ehewz.baihongyu.com/

你可能感兴趣的文章
MySQL 中锁的面试题总结
查看>>
MySQL 中随机抽样:order by rand limit 的替代方案
查看>>
MySQL 为什么需要两阶段提交?
查看>>
mysql 为某个字段的值加前缀、去掉前缀
查看>>
mysql 主从
查看>>
mysql 主从 lock_mysql 主从同步权限mysql 行锁的实现
查看>>
mysql 主从互备份_mysql互为主从实战设置详解及自动化备份(Centos7.2)
查看>>
mysql 主从关系切换
查看>>
mysql 主键重复则覆盖_数据库主键不能重复
查看>>
Mysql 优化 or
查看>>
mysql 优化器 key_mysql – 选择*和查询优化器
查看>>
MySQL 优化:Explain 执行计划详解
查看>>
Mysql 会导致锁表的语法
查看>>
mysql 使用sql文件恢复数据库
查看>>
mysql 修改默认字符集为utf8
查看>>
Mysql 共享锁
查看>>
MySQL 内核深度优化
查看>>
mysql 内连接、自然连接、外连接的区别
查看>>
mysql 写入慢优化
查看>>
mysql 分组统计SQL语句
查看>>