FatMouse and Cheese

FatMouse and Cheese

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

 

/*
题意：给你一个n*n的矩阵，每个点都有一定数量的食物，出发点在(0,0)，每次可以走最多k步，下一个单元格的食物数量
    一定要比当前单元格的数量多，问你最多能吃到的食物数量

初步思路：记忆化搜索
*/
#include <bits/stdc++.h>
#define N 110
using namespace std;
int n,k;
int mapn[N][N];
int dp[N][N];//dp[i][j]用于保存到达i j之后能吃到的最大食物数量
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
bool vis[N][N];
bool ok(int x,int y,int val){
    if(x<0||x>=n||y<0||y>=n||vis[x][y]||mapn[x][y]<=val) return false;
    return true;
}
int dfs(int x,int y){//三种 状态坐标，第几步
    if(dp[x][y]!=-1) //遇到合适的状态直接返回就行了
        return dp[x][y];
    int cur=0;
    for(int i=0;i<4;i++){
        int fx=x,fy=y;
        for(int j=1;j<=k;j++){
            fx+=dir[i][0];
            fy+=dir[i][1];
            if(ok(fx,fy,mapn[x][y])==false) continue;

            vis[fx][fy]=true;

            cur=max(cur,dfs(fx,fy));

            vis[fx][fy]=false; //一条路走完了，将标记清理干净
        }
    }
    return dp[x][y]=cur+mapn[x][y];
}
void init(){
    memset(dp,-1,sizeof dp);
    memset(vis,false,sizeof vis);
}
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&k)!=EOF&&(n!=-1&&k!=-1)){
        init();
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                scanf("%d",&mapn[i][j]);
            }
        }//处理输入
        dfs(0,0);
        // for(int i=0;i<n;i++){
        //     for(int j=0;j<n;j++){
        //         cout<<dp[i][j]<<" ";
        //     }
        //     cout<<endl;
        // }
        printf("%d\n",dp[0][0]);
    }
    return 0;
}