FatMouse and Cheese
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9499 Accepted Submission(s): 4007
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.
//普通的搜索会超时，记忆化搜索，当搜到某点的权值和在前面的搜索中已经算过了就直接返回。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int inf=0x7fffffff;
int f[][],mp[][];
int dir[][]={,,-,,,,,-};
int n,k;
int dfs(int x,int y)
{
if(!f[x][y]){
int tmp=;
for(int j=;j<=k;j++){
for(int i=;i<;i++){
int xx=x+dir[i][]*j,yy=y+dir[i][]*j;
if(xx<||xx>=n||yy<||yy>=n) continue;
if(mp[xx][yy]<=mp[x][y]) continue;
tmp=max(tmp,dfs(xx,yy));
}
}
f[x][y]=tmp+mp[x][y];
}
return f[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",&mp[i][j]);
}
}
memset(f,,sizeof(f));
int ans=dfs(,);
printf("%d\n",ans);
}
return ;
}