HDU1078 FatMouse and Cheese(DFS+DP) 2016-07-24 14:05 70人阅读 评论(0) 收藏
FatMouse and Cheese
Problem Description
cheese in a hole. Now he's going to enjoy his favorite food.
FatMouse begins by standing at location (0,0). He eats up the cheese where he stands and then runs either horizontally or vertically to another location. The problem is that there is a super Cat named Top Killer sitting near his hole, so each time he can run
at most k locations to get into the hole before being caught by Top Killer. What is worse -- after eating up the cheese at one location, FatMouse gets fatter. So in order to gain enough energy for his next run, he has to run to a location which have more blocks
of cheese than those that were at the current hole.
Given n, k, and the number of blocks of cheese at each grid location, compute the maximum amount of cheese FatMouse can eat before being unable to move.
Input
a line containing two integers between 1 and 100: n and k
n lines, each with n numbers: the first line contains the number of blocks of cheese at locations (0,0) (0,1) ... (0,n-1); the next line contains the number of blocks of cheese at locations (1,0), (1,1), ... (1,n-1), and so on.
The input ends with a pair of -1's.
Output
Sample Input
3 1
1 2 5
10 11 6
12 12 7
-1 -1
Sample Output
37
——————————————————————————————————————————————————————
#include <iostream>
#include <cstring>
using namespace std; int dir[4][2] = { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } };
int mp[105][105];
int dp[105][105];
int m, k; bool cheak(int i, int j)
{
if (i < 0 || i >= m || j < 0 || j >= m)
return 0;
else
return 1;
} int dfs(int x, int y)
{
if (dp[x][y] > 0)
return dp[x][y];
int mx = 0;
int xx, yy;
for (int i = 0; i < 4;i++)
for (int j = 1; j <= k; j++)
{
xx = x + dir[i][0] * j;
yy = y + dir[i][1] * j;
if (cheak(xx, yy) && mp[xx][yy]>mp[x][y])
{
int t = dfs(xx, yy);
if (t>mx)
mx = t;
}
}
dp[x][y] = mp[x][y] + mx;
return dp[x][y];
} int main()
{
while (scanf("%d%d", &m, &k) && (m != -1 || k != -1))
{
for (int i = 0; i < m;i++)
for (int j = 0; j < m; j++)
{
scanf("%d", &mp[i][j]);
}
memset(dp, 0, sizeof(dp));
int ans=dfs(0, 0);
printf("%d\n", ans); } return 0; }
HDU1078 FatMouse and Cheese(DFS+DP) 2016-07-24 14:05 70人阅读 评论(0) 收藏的更多相关文章
- A Knight's Journey 分类: dfs 2015-05-03 14:51 23人阅读 评论(0) 收藏
A Knight’s Journey Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 34085 Accepted: 11621 ...
- Hdu1016 Prime Ring Problem(DFS) 2016-05-06 14:27 329人阅读 评论(0) 收藏
Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- HDU1426 Sudoku Killer(DFS暴力) 2016-07-24 14:56 65人阅读 评论(0) 收藏
Sudoku Killer Problem Description 自从2006年3月10日至11日的首届数独世界锦标赛以后,数独这项游戏越来越受到人们的喜爱和重视. 据说,在2008北京奥运会上,会 ...
- HDU1258 Sum It Up(DFS) 2016-07-24 14:32 57人阅读 评论(0) 收藏
Sum It Up Problem Description Given a specified total t and a list of n integers, find all distinct ...
- leetcode N-Queens/N-Queens II, backtracking, hdu 2553 count N-Queens, dfs 分类: leetcode hdoj 2015-07-09 02:07 102人阅读 评论(0) 收藏
for the backtracking part, thanks to the video of stanford cs106b lecture 10 by Julie Zelenski for t ...
- Hdu 1009 FatMouse' Trade 分类: Translation Mode 2014-08-04 14:07 74人阅读 评论(0) 收藏
FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- hdu 1231, dp ,maximum consecutive sum of integers, find the boundaries, possibly all negative, C++ 分类: hdoj 2015-07-12 03:24 87人阅读 评论(0) 收藏
the algorithm of three version below is essentially the same, namely, Kadane's algorithm, which is o ...
- A Knight's Journey 分类: POJ 搜索 2015-08-08 07:32 2人阅读 评论(0) 收藏
A Knight's Journey Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 35564 Accepted: 12119 ...
- HDU1506(单调栈或者DP) 分类: 数据结构 2015-07-07 23:23 2人阅读 评论(0) 收藏
Largest Rectangle in a Histogram Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 ...
随机推荐
- Haskell语言学习笔记(52)正则表达式
Text.Regex.PCRE.Heavy $ brew install pcre $ cabal install pcre-heavy Installed pcre-heavy-1.0.0.2 Pr ...
- asp.net cors solution
I have a simple actionmethod, that returns some json. It runs on ajax.example.com. I need to access ...
- git 拉取某个分支到本地
git 拉取其实只需要 git fetch origin xxx. git pull origin xxx即可
- 树莓派Zero W GPIO控制
作者:陈拓 chentuo@ms.xab.ac.cn 2018.06.09/2018.07.05 0. 概述 本文介绍树莓派 Zero W的GPIO控制,并用LED看效果. 0.1 树莓派GPIO编 ...
- vim使用方法:
vim使用方法: 模式: 编辑模式.未编辑模式.命令行模式 i 插入形式进入编辑模式 a 增加 o 下行编辑 O 上行插入 : 进入命令行模式 esc 退出编辑模式 wq 保存文件 yy 复制 p 粘 ...
- Electron Browser加载iframe(webview src属性)
browser或者webcontents 的高度与宽度比例对webview中src的页面结构也是有一定影响的
- 为什么JAVA要提供 wait/notify 机制?是为了避免轮询带来的性能损失
wait/notify 机制是为了避免轮询带来的性能损失. 为了说清道理,我们用“图书馆借书”这个经典例子来作解释. 一本书同时只能借给一个人.现在有一本书,图书馆已经把这本书借了张三. 在简单的s ...
- Hook钩子编程
钩子(Hook),是Windows消息处理机制的一个平台,钩子实际上是一个处理消息的程序段,通过系统调用,把它挂入系统,以监视指定窗口的某种消息.每当特定的消息发出,在没有到达目的窗口前,钩子程序就先 ...
- Petya and Graph(最小割,最大权闭合子图)
Petya and Graph http://codeforces.com/contest/1082/problem/G time limit per test 2 seconds memory li ...
- Hibernate查询方式(补)
-----------------siwuxie095 Hibernate 查询方式 1.对象导航查询 根据已经加载的对 ...