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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题意：老鼠吃奶酪，从（0，0）出发每步只能垂直或水平走1-k格，且下一步的奶酪数要比现在的大，输出走过奶酪数和的最大值。

思路：记忆化搜索，dp[]记录的是最终从这走能走到的最大和，所以是从最后往回累加 。

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1 #include
     
       2 #include
      
        3 #include
       
         4 using namespace std; 5 int n,k; 6 int map[105][105]; 7 int dp[105][105]; 8 int dir[4][2]={
   {-1,0},{
   0,1},{
   1,0},{
   0,-1}}; 9 10 int dfs(int x,int y){11     int ans=0;12     if(!dp[x][y]){
   //如果已经有值，则说明已经找到了该点的最大和直接用就行。    13         for(int j=1;j<=k;j++){    14            for(int i=0;i<4;i++){15                 int xx=x+dir[i][0]*j;//水平跑 16                 int yy=y+dir[i][1]*j;//垂直跑 17                 if(xx<0||yy<0||xx>=n||yy>=n||map[xx][yy]<=map[x][y]) continue;             18                 ans=max(ans,dfs(xx,yy));    19             }    20         }21         dp[x][y]=ans+map[x][y]; 22     } 23     return dp[x][y];24 }25 26 int main(){27     while(scanf("%d%d",&n,&k)){28         if(n==-1&&k==-1) break;29         30         for(int i=0;i