FatMouse and CheeseI - I

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. 

3 1
1 2 5
10 11 6
12 12 7
-1 -1

37

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<queue>
 4 #include<set>
 5 #include<iostream>
 6 #include<map>
 7 #include<stack>
 8 #include<cmath>
 9 #include<algorithm>
10 #define ll long long
11 using namespace std;
12 int n,k;
13 int f[4][2]={1,0,0,1,-1,0,0,-1};
14 int dp[1000][1000],e[1000][1000];
15 int dfs(int x,int y)
16 {
17     if(dp[x][y]) return dp[x][y];
18     for(int i=0;i<4;i++)
19     {
20         for(int j=1;j<=k;j++)
21         {
22             int tx=x+j*f[i][0];
23             int ty=y+j*f[i][1];
24             if(tx>=0&&ty>=0&&tx<n&&ty<n&&e[tx][ty]>e[x][y])
25             {
26                 dp[x][y]=max(dfs(tx,ty)+e[tx][ty],dp[x][y]);
27             }
28         }
29     }
30     return dp[x][y];
31 }
32 int main()
33 {
34     while(~scanf("%d%d",&n,&k)&&(n+k)!=-2)
35     {
36         memset(dp,0,sizeof(dp));
37         for(int i=0;i<n;i++)
38             for(int j=0;j<n;j++)
39                 scanf("%d",&e[i][j]);
40         printf("%d\n",dfs(0,0)+e[0][0]);
41     }
42 }