FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14888    Accepted Submission(s): 6300

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

題意：n * n的正方形格子（每個格子均放了乳酪），老鼠從(0,0)開始，每次最多移動k步，可以選擇上下左右四個方向移動，下一個移動點乳酪塊數量必須要大於當前點。

可以想到是搜尋，但是沒有用陣列去儲存每次的最大數量就會dfs超時。

 記憶化搜尋入門

AC Code: 

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<queue>
using namespace std;
static const int dir[4][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
int n, k;
int dp[105][105], vv[105][105];
inline int read(){int x = 0, f = 1;char ch = getchar();
    while(ch > '9' || ch < '0') {if(ch == '-') f = -1;ch = getchar();}
    while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}
bool is_bound(int x, int y){
    return (x >= 0 && x < n && y >= 0 && y < n);
}

int dfs(int x, int y){
    if(dp[x][y]) return dp[x][y];
    int res = 0;
    for(int t = 1; t <= k; t++){
        for(int i = 0; i < 4; i++){
            int fx = x + dir[i][0] * t;
            int fy = y + dir[i][1] * t;
            if(!is_bound(fx, fy)) continue;
            if(vv[fx][fy] > vv[x][y])
                res = max(res, dfs(fx, fy));
        }
    }
    dp[x][y] = res + vv[x][y];
    return dp[x][y];
}
int main(){
    while(scanf("%d%d", &n, &k) != EOF){
        memset(dp, 0, sizeof(dp));
        if(n == -1 && k == -1) break;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                scanf("%d", &vv[i][j]);
            }
        }
        printf("%d\n", dfs(0, 0));
    }
    return 0;
}