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 問題描述 在第一行輸入n和k。表示有n*n個老鼠洞，k表示每一次老鼠可以移動多長距離。 接下來的n行n列存儲每一個老鼠洞中存儲的奶酪的多少。每一次胖老鼠只能移動到存有更多奶酪的老鼠洞中。問老鼠從（0，0）點開始最多可以吃到多少奶酪。 問題分析 假設(shè)有一個函數(shù)int find(int x,int y);返回胖老鼠從當(dāng)前節(jié)點開始最多能吃到多少奶酪。 那么我們只需要遍歷從當(dāng)前節(jié)點開始能夠到達(dá)的點(x,y)計算find(x,y)取最大值加上當(dāng)前節(jié)點的奶酪數(shù)量即為最大值。 注：要記得存儲當(dāng)前計算過的值避免重復(fù)計算 代碼實現(xiàn)