Fatmouse and cheese

肥鼠在n×n的网格中储存了奶酪,从(0,0)开始,每次最多跑k步且必须前往奶酪多的位置。目标是找出在被超级猫Top Killer抓到前,能吃到的最大奶酪数量。输入包含n、k及每个位置的奶酪数,输出最大收集量。问题涉及到深度优先搜索策略。" 119987996,213016,高斯求和公式解析与C语言实现,"['数学', '编程', '算法', 'C语言', '计算机科学']

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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.(问题描述;肥鼠在一个城市里储存了一些奶酪。这个城市可以被认为是一个n维的正方形网格:每个网格位置都标有(p,q),其中0 <= p < n,0 <= q < n。在每个网格位置,肥鼠都在一个洞里藏了0到100块奶酪。现在他要享受他最喜欢的食物了。)

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.(最后一个鼠标从位置(0,0)开始。他在原地吃完奶酪,然后水平或垂直地跑向另一个地方。问题是有一只名叫托普·黑仔的超级猫坐在他的洞附近,所以每次他都可以在被托普·黑仔抓住之前跑到最多k个位置进入洞内。更糟糕的是——在一个地方吃完奶酪后,肥鼠变得更胖了。因此,为了获得足够的能量进行下一次跑步,他必须跑到一块奶酪块比当前洞多的地方。)

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.(给定n、k和每个网格位置的奶酪块数量,计算FatMouse在无法移动之前可以吃的最大奶酪量。)

Input
There are several test cases. Each test case consists of(输入;有几个测试案例。每个测试用例包括)

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.(包含1和100之间的两个整数的线:n和k;n行,每一行有n个数字:第一行包含位置(0,0) (0,1)处奶酪块的数量…(0,n-1);下一行包含位置(1,0)、(1,1)处奶酪块的数量,…(1,n-1)等等。输入以一对-1结尾。)

Output
For each test case output in a line the single integer giving the number of blocks of cheese collected.(输出;对于一行中的每个测试用例输出,给出收集的奶酪块数量的单个整数)

#include <iostream>
#include <cstdio>
#include <cstring>
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值