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

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思路：

特别记得还在大一下学期军训（5月份）的时候，第一次接触到dfs和bfs的相关题目，当时做了用了好长时间才勉强能做出几道，现在这个题看了下题解基本就掌握套路了

对于老鼠所到达的每个位置，它下一步可能到达的位置有上下左右4个方向分别走w（1=<w<=k）步，方向用t数组来记录

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#include 
      
       #include 
       
        #include 
        
          using namespace std;int n,k;int map[107][107];int dp[107][107];int t[4][2]={
    1,0,-1,0,0,-1,0,1};int max(int a,int b) {    return a>b?a:b;}int dfs(int x,int y){    int maxn = 0;    int ans;    if(dp[x][y] != -1) return dp[x][y];    else {        for(int i = 1;i <= k;i++)             for(int j = 0;j < 4;j++)             {                int new_x = x+t[j][0]*i;                int new_y = y+t[j][1]*i;                if(new_x>=0&&new_x
         
          =0&&new_y
          
           map[x][y])                {                    ans = dfs(new_x,new_y);                    maxn = max(ans,maxn);                }            }        return dp[x][y] = maxn+map[x][y];    }}int main(){    while(scanf("%d%d",&n,&k))    {        if(n==-1 && k==-1) break;        for(int i = 0;i < n;i++)             for(int j = 0;j