终于开始写dp了,还很不熟练

It is a little known fact that cows love apples. Farmer John has two apple trees (which are conveniently numbered 1 and 2) in his field, each full of apples. Bessie cannot reach the apples when they are on the tree, so she must wait for them to fall. However, she must catch them in the air since the apples bruise when they hit the ground (and no one wants to eat bruised apples). Bessie is a quick eater, so an apple she does catch is eaten in just a few seconds.

Each minute, one of the two apple trees drops an apple. Bessie, having much practice, can catch an apple if she is standing under a tree from which one falls. While Bessie can walk between the two trees quickly (in much less than a minute), she can stand under only one tree at any time. Moreover, cows do not get a lot of exercise, so she is not willing to walk back and forth between the trees endlessly (and thus misses some apples).

Apples fall (one each minute) for T (1 <= T <= 1,000) minutes. Bessie is willing to walk back and forth at most W (1 <= W <= 30) times. Given which tree will drop an apple each minute, determine the maximum number of apples which Bessie can catch. Bessie starts at tree 1.

Input

* Line 1: Two space separated integers: T and W

* Lines 2..T+1: 1 or 2: the tree that will drop an apple each minute.

Output

* Line 1: The maximum number of apples Bessie can catch without walking more than W times.

Sample Input

7 2
2
1
1
2
2
1
1

Sample Output

6

Hint

INPUT DETAILS:

Seven apples fall - one from tree 2, then two in a row from tree 1, then two in a row from tree 2, then two in a row from tree 1. Bessie is willing to walk from one tree to the other twice.

OUTPUT DETAILS:

Bessie can catch six apples by staying under tree 1 until the first two have dropped, then moving to tree 2 for the next two, then returning back to tree 1 for the final two.

 
分析:状态:dp[i][j]表示在第i分钟时,已经移动了j次后得到的苹果数量。
状态转移方程:dp[i][j] = max(dp[i-1][j], dp[i-1][j-1]),然后判断当前是否在第i分钟掉苹果的那颗树下,是的话,dp[i][j]++。
对状态转移方程的解释如下:第i分钟能得到的苹果数量,等于在第i-1分钟时,在树1和树2下得到苹果的最大值。j为偶数则在树1下面,奇数则在树2下面。

dp 动态规划 之C - Apple Catching 简单基础的更多相关文章

  1. poj 2385 Apple Catching 基础dp

    Apple Catching   Description It is a little known fact that cows love apples. Farmer John has two ap ...

  2. poj2385 Apple Catching (线性dp)

    题目传送门 Apple Catching Apple Catching Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 154 ...

  3. Apple Catching(dp)

    Apple Catching Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9831   Accepted: 4779 De ...

  4. BZOJ 3384: [Usaco2004 Nov]Apple Catching 接苹果( dp )

    dp dp( x , k ) = max( dp( x - 1 , k - 1 ) + *** , dp( x - 1 , k ) + *** ) *** = 0 or 1 ,根据情况 (BZOJ 1 ...

  5. 【POJ】2385 Apple Catching(dp)

    Apple Catching Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13447   Accepted: 6549 D ...

  6. 【POJ - 2385】Apple Catching(动态规划)

    Apple Catching 直接翻译了 Descriptions 有两棵APP树,编号为1,2.每一秒,这两棵APP树中的其中一棵会掉一个APP.每一秒,你可以选择在当前APP树下接APP,或者迅速 ...

  7. POJ 2385 Apple Catching【DP】

    题意:2棵苹果树在T分钟内每分钟随机由某一棵苹果树掉下一个苹果,奶牛站在树#1下等着吃苹果,它最多愿意移动W次,问它最多能吃到几个苹果.思路:不妨按时间来思考,一给定时刻i,转移次数已知为j, 则它只 ...

  8. Day 5 笔记 dp动态规划

    Day 5 笔记 dp动态规划 一.动态规划的基本思路 就是用一些子状态来算出全局状态. 特点: 无后效性--狗熊掰棒子,所以滚动什么的最好了 可以分解性--每个大的状态可以分解成较小的步骤完成 dp ...

  9. (转)dp动态规划分类详解

    dp动态规划分类详解 转自:http://blog.csdn.NET/cc_again/article/details/25866971 动态规划一直是ACM竞赛中的重点,同时又是难点,因为该算法时间 ...

随机推荐

  1. Java 在 CMD 环境下编译

    1. 未引用第三方 Jar 包类(该 Java 类默认无包名) # 定位到类存放地址 cd E:\tsgg # 编译命令 javac Test.java # 执行命令 java Test 2. 引用第 ...

  2. 一张图读懂PBN飞越转弯衔接TF/CF航段计算

    在PBN旁切转弯的基础上,再来看飞越转弯接TF(或CF)航段,保护区结构上有些相似,只是转弯拐角处的保护区边界有“简化”,其余部分是相近的. FlyOver接TF段的标称航迹有一个飞越之后转弯切入航迹 ...

  3. windows环境下pycharm如何设置Linux编码

    最近写代码一直在windows环境下,写完之后再传到Linux端就会出现代码格式错乱. 解决办法: 在windows端的pycharm代码格式设置为unix and os及可以解决这个问题. 如果你要 ...

  4. 再会Java

    作者曾写过一段时间Java, 时间一长也就忘得差不多了. 现在重新学习一个, 故而只是提要式的记录. Java是静态强类型语言, 运行于Java虚拟机(Java Virtual Machine, JV ...

  5. jQuery学习(1)猜数字游戏

      jQuery是一个快捷.小型且特征丰富的JavaScript库.它使得HTML文档遍历及操作,事件处理,动画,Ajax等更简洁方便.它通过调用一个简单易用的API,就能在各种浏览器中使用.由于jQ ...

  6. T-SQL :TOP和OFFSET-FETCH筛选 (五)

    通过were和having条件可以对数据进行筛选,那么如何通过排序对数据进行筛选呢? 1.TOP筛选 用于限制查询返回行数或者行数的百分比. 例如 我们对订单表筛选最近产生的订单5条 ) orderi ...

  7. [android] 手机卫士绑定sim卡

    更新: 收不到启动广播,查看知乎,好像是说高版本的系统都禁止了 还可以通过adb发送开机广播 adb shell am broadcast -a android.intent.action.BOOT_ ...

  8. (3)Jquery1.8.3快速入门_jquery对象dom对象转换

    1.Jquery 对象 dom对象的转化使用: 1.1.jquery 对象: 通过$()包装DOM对象后产生的对象. 1.2.jquery对象是Jquery独有的 ,可以使用jquery中的方法. 1 ...

  9. Java学习笔记之——冒泡排序

    冒泡排序:解决数组的排序问题,比如从大到小或者从小到大 原理:两两比较 案例:

  10. Layui tree 下拉菜单树

    1.效果: 2.html  代码: <!DOCTYPE html> <html> <head> <meta charset="utf-8" ...