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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 11 2 510 11 612 12 7-1 -1

Sample Output

37

题意：n,k,n*n型的数据矩阵，每一个位置都有一个数，到一个地方就可以获得那个位置上的数（要到的那个地方上的数必须大于此时你所在位置上的数），问从（0,0）处出发，可以向四个方向走，一个方向可以选择走（1~k）步，问题是最多可以拿到多少？

思路：dp加上搜索回溯，从（0,0）点开始扩散，直到扩散到最大数，然后就开始回溯，此时加上dp,记忆化搜索，减少时间的复杂度，直到回溯到（0,0）点，即为最大值。

#include <stdio.h>#include <string.h>#define max(a,b)((a)>(b)?(a):(b))int n,m;int map[110][110];int dp[110][110];int dis[4][2]={0,1,1,0,0,-1,-1,0};int dfs(int x,int y){if(dp[x][y])return dp[x][y];int max1=0;for(int i=0;i<4;i++)for(int j=1;j<=m;j++){int tx=x+dis[i][0]*j;int ty=y+dis[i][1]*j;if(tx<0||ty<0||tx>=n||ty>=n)continue;if(map[tx][ty]>map[x][y])max1=max(max1,dfs(tx,ty));}return dp[x][y]=max1+map[x][y];}int main(){  while(~scanf("%d %d",&n,&m)&&(n!=-1||m!=-1))  {  for(int i=0;i<n;i++)   for(int j=0;j<n;j++)     scanf("%d",&map[i][j]);  memset(dp,0,sizeof(dp));  printf("%d\n",dfs(0,0));  }  }