FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description

FatMouse has stored some cheese in a city. The city can be considered as a square grid of dimension n: each grid location is labelled (p,q) where 0 <= p < n and 0 <= q < n. At each grid location Fatmouse has hid between 0 and 100 blocks of 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

There are several test cases. Each test case consists of

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

For each test case output in a line the single integer giving the number of blocks of cheese collected.

Sample Input

3 1

1 2 5

10 11 6

12 12 7

-1 -1

Sample Output

37

题意:给一个图,从(0,0)开始走,只能走上向左右,所去的位置上的数要大于当前位置,每到一个位置可以将这个位置的数累加到自身,问最多能取多大的数。

题解:DP题目,记忆+DFS。注意是多组输入,(-1,-1)停止。

#include <iostream>

using namespace std;

int n,dp[105][105],s[105][105],k;

int Next[4][2] = {1,0,-1,0,0,-1,0,1};

int juedge(int x,int y)
{
if(x>=0&&y>=0&&x<n&&y<n)
return 1;
return 0;
} int DFS(int x,int y)
{
if(dp[x][y])
return dp[x][y];
int i,j,dx,dy,MAX;
MAX = 0;
for(i=1;i<=k;i++)
{
for(j=0;j<4;j++)
{
dx = x + Next[j][0] * i;
dy = y + Next[j][1] * i;
if(juedge(dx,dy)&&s[dx][dy]>s[x][y])
{
int a = DFS(dx,dy);
if(a>MAX)
MAX = a;
}
}
}
dp[x][y] = MAX + s[x][y];
return dp[x][y];
} int main()
{
int i,j;
while(cin>>n>>k)
{
if(n==-1&&k==-1)
break;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
dp[i][j] = 0;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
cin>>s[i][j];
DFS(0,0);
cout<<dp[0][0]<<endl;
}
return 0;
}

HUD-1708_FatMouse and Cheese的更多相关文章

  1. 如何用Unity GUI制作HUD

    若知其所以然,自然知其然. HUD是指平视显示器,就是套在脸上,和你的眼睛固定在一起,HUD的意思就是界面咯,一般我们说HUD特指把3D空间中的界面的某些信息(比如血条,伤害之类)的贴在界面上,对应3 ...

  2. xamarin UWP平台下 HUD 自定义弹窗

    在我的上一篇博客中我写了一个在xamarin的UWP平台下的自定义弹窗控件.在上篇文章中介绍了一种弹窗的写法,但在实际应用中发现了该方法的不足: 1.当弹窗出现后,我们拖动整个窗口大小的时候,弹窗的窗 ...

  3. xamarin UWP设置HUD加载功能

    使用xamarin开发的时候经常用到加载HUD功能,就是我们常见的一个加载中的动作,Android 下使用 AndHUD , iOS 下使用 BTProgressHUD, 这两个在在 NuGet 上都 ...

  4. CF 371B Fox Dividing Cheese[数论]

    B. Fox Dividing Cheese time limit per test 1 second memory limit per test 256 megabytes input standa ...

  5. hdu 1078 FatMouse and Cheese

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...

  6. OSG中的HUD

    OSG中的HUD 所谓HUD节点,说白了就是无论三维场景中的内容怎么改变,它都能在屏幕上固定位置显示的节点. 实现要点: 关闭光照,不受场景光照影响,所有内容以同一亮度显示 关闭深度测试 调整渲染顺序 ...

  7. CREATE A ENERGY / HEALTH BAR HUD

    Now then, let's get started. 1. Open the Play scene which you had created in the previous post. If y ...

  8. iOS之UI--指示器HUD的创建和设置

    指示器的创建和设置 渐变动画 描述: 使用label就能制作指示器,原理:就是让label以动画的形式慢慢显示和消失 最好是半透明的 指示器有时候也被称为:HUD,遮盖,蒙版 思路步骤: 1.先在st ...

  9. NGUI:HUD Text(头顶伤害漂浮文字)

    HUD Text 很早之前就有闻于NGUI中的HUD Text插件,今天得以尝试,看了会儿官方的文档,楞是没给看明白,官方的ReadMe.txt写的使用方法如下: 官网Usage 1. Attach ...

  10. HDU 1078 FatMouse and Cheese(记忆化搜索)

    FatMouse and Cheese Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Othe ...

随机推荐

  1. Django项目:CRM(客户关系管理系统)--34--26PerfectCRM实现King_admin自定义排序

    ordering = ['-qq'] #自定义排序,默认'-id' #base_admin.py # ————————24PerfectCRM实现King_admin自定义操作数据———————— f ...

  2. body-parser与querystring与multer的区别

    body-parser express中间件 body-parser是用来解析http请求体的,是express默认使用的中间件之一. (只解析post的普通数据请求,无法解析post文件请求) 使用 ...

  3. CodeChef--EQUAKE

    题目链接 Earthquake in Bytetown! Situation is getting out of control! All buildings in Bytetown stand on ...

  4. C#中使用设置(Settings.settings) Properties.Settings.Default

    应用程序及用户设置 在设计时创建新设置的步骤 在“Solution Explorer”(解决方案资源管理器)中,展开项目的“Properties”(属性)节点. 在“Solution Explorer ...

  5. VMWare 下 Ubuntu 18.04 的文件共享

    突然某天发现 /mnt/hgfs 下共享的文件夹没了... apt-get install open-vm-tools mkdir /mnt/hgfs vmhgfs-fuse .host:/ /mnt ...

  6. jquery 获取图片宽高为0的问题

    原理:页面加载完了,图片不一定加载完了. $(function(){ $("img").on("load",function(){ //核心 var w = $ ...

  7. java swing多线程

    比如一个爬虫 在界面上显示当前时间,每秒都刷新一次用来判断软件是不是卡死 在爬取程序运行的时候,界面可能会卡死 那这就要把爬取程序放在另一个线程里边 同时,也可以把rtc放在另一个线程里边 具体代码, ...

  8. bzoj3064/洛谷P4314 CPU监控【线段树】

    好,长草博客被催更了[?] 我感觉这题完全可以当作线段树3 线段树2考加法和乘法标记的下放顺序,这道题更丧心病狂[?] 很多人可能跟我一样,刚看到这道题秒出思路:打一个当前最大值一个历史最大值不就完事 ...

  9. 验证python中函数传参是引用传递

    定义: 值传递(pass by value)是指在调用函数时将实际参数复制一份传递到函数中,这样在函数中如果对参数进行修改,将不会影响到实际参数. 引用传递(pass by reference)是指在 ...

  10. 稳定性专题 | StackOverFlowError 常见原因及解决方法

    导读 『StabilityGuide』是阿里多位阿里技术工程师共同发起的稳定性领域的知识库开源项目,涵盖性能压测.故障演练.JVM.应用容器.服务框架.流量调度.监控.诊断等多个技术领域,以更结构化的 ...