Apple Catching
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7076 | Accepted: 3453 |
Description
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.
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.
* 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.
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.
题意:
给出 T(1 ~ 1000),W(1 ~ 30),代表有 T 秒,允许走的步数是 W。一共有两棵树,1 和 2 ,每秒中都会有其中一棵树掉苹果下来。后给出这 7s 内掉的分别是哪棵树,问在 W 秒内,最多能接到几个苹果。输出这个数。
思路:
DP。设 dp [ i ] [ j ] [ k ] 代表 i 秒时候的 j 步内于 k 号数能接到的最大苹果数。所以有两种情况:
dp [ i ] [ j ] [ 1 ] = max ( dp [ i - 1 ] [ j ] [ 1 ] , dp [ i - 1 ] [ j - 1 ] [ 2 ] ) + tree1 [ i ] ;
dp [ i ] [ j ] [ 2 ] = max ( dp [ i - 1 ] [ j ] [ 2 ] , dp [ i - 1 ] [ j - 1 ] [ 1 ] ) + tree2 [ i ] ;
初始化要初始 j = 0 即步数为 0 时候的。
AC:
#include <cstdio> #include <algorithm> #include <cstring> #include <iostream> using namespace std; const int MAX = 1005; int tree1[MAX], tree2[MAX]; int dp[MAX][35][5]; int main () { int t, w; scanf("%d%d", &t, &w); memset(tree1, 0, sizeof(tree1)); memset(tree2, 0, sizeof(tree2)); for (int i = 1; i <= t; ++i) { int ans; scanf("%d", &ans); if (ans == 1) tree1[i] = 1; else tree2[i] = 1; } for (int i = 1; i <= t; ++i) { dp[i][0][1] = dp[i - 1][0][1] + tree1[i]; dp[i][0][2] = 0; } for (int i = 1; i <= t; ++i) { for (int j = 1; j <= w; ++j) { dp[i][j][1] = max(dp[i - 1][j - 1][2], dp[i - 1][j][1]) + tree1[i]; dp[i][j][2] = max(dp[i - 1][j - 1][1], dp[i - 1][j][2]) + tree2[i]; } } printf("%d\n", max(dp[t][w][1], dp[t][w][2])); return 0; }
相关推荐
Packet filtering - Catching the cool packets.rar
Catching header messages in a CListView捕捉CListView的头消息(2KB)
C++ 用编程方式实现视频,照相捕捉功能。
This book is designed to support anyone who wants to work in networking with a focus on troubleshooting, optimization and security.
基于matlab实现黑人抬棺的音乐,代码很短较为简单,适合初学者。简单的代码,就可以实现功能,可以用于学习。
一个彷b站醒目留言的控件,复制即可用,欢迎下载。原博客地址:https://blog.csdn.net/Sure_Min/article/details/107141315
Catching popular prefixes at AS border routers with a prediction based method
语言:English 嵌入Cera以在某些%的页面上查找
2008年硕士研究生考试英语真题详解 完型填空 1、答案:B 解析:本题测试语义逻辑衔接。 “ selected” 意为 “挑选”; “prepared”意为 “准备”; “obliged”意为“迫使,责成”;“pleased”意为“高兴地...
嵌入一个Cera来找到一些页面 支持语言:English
语言:English,English (UK),English (United States) 钓鱼冲突:捉鱼游戏,现在可在Chrome浏览器上使用 与《钓鱼冲突:捉鱼游戏》中的电子游戏史上最著名的英雄一起玩。 ... -在左上角,单击设置以根据需要自定义所有...
C39-调试-第1部分
Create a user interface that is eye-catching and stands apart from the crowd Maximize your use of typographic elements for style and readability Perfect entry views and display large amounts of data...
An app for catching up on things. https://medium.com/@sweers/catching-up-on-catchup-introduction-7581c099f4bc Motivations There's a lot of services I like reading up on throughout the day. Most of ...
iOS 10 SDK Development: Creating iPhone and iPad ... Whether you're new to Swift or just catching up on iOS' latest features, iOS 10 SDK Development will help you master the language and the platform.
异常捕获---BOXES 展示了如何捕获异常以使错误易于阅读! 补丁 0.3
Juego con JavaScript Juego desarrollado配置了JavaScript,HTML和CSS... Ayuda a Scrat a atrapar su alimento最喜欢的人:“ nueces! 通过皮拉尔·加西亚(PilarGarcía)。 人因工厂,Distrtal大学FJDC,2017年。
You'll not only learn how to design for Apple's devices, you'll also master the iPhone SDK tools -- including Interface Builder, Xcode, and Objective-C programming principles -- to make your app ...