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.
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.
1 2 5
10 11 6
12 12 7
-1 -1
思路：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std; int n,k;
int map[][];
int dp[][];
int t[][]={,,-,,,-,,};
int max(int a,int b) {
return a>b?a:b;
} int dfs(int x,int y)
{
int maxn = ;
int ans;
if(dp[x][y] != -) return dp[x][y];
else {
for(int i = ;i <= k;i++)
for(int j = ;j < ;j++)
{
int new_x = x+t[j][]*i;
int new_y = y+t[j][]*i;
if(new_x>=&&new_x<n&&new_y>=&&new_y<n&&map[new_x][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==- && k==-) break;
for(int i = ;i < n;i++)
for(int j = ;j <n;j++)
scanf("%d",&map[i][j]);
memset(dp,-,sizeof(dp));
printf("%d\n",dfs(,));
}
return ;
}