HDOJ1078(FatMouse and Cheese)

FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 831    Accepted Submission(s): 253

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

 

 

/**/ /*1.each time he can run at most k locations to get into the hole 
  2.he has to run to a location which have more blocks of cheese 
  than those that were at the current hole;
  */
//08:46:39 Accepted 1078 62MS 284K 1567 B C++ Xredman 
#include <iostream>
#include <utility>
using namespace std;

const int N = 102;


int graph[N][N];
int collect[N][N];
int n, k, ans;
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

bool isBound(int x, int y)
{//对坐标是否越界进行判断
    if(x < 0 || y < 0)
        return false;
    if(x >= n || y >= n)
        return false;
    return true;
}

int memory(pair<int, int> p )
{
    int i, j, maxb = 0, tmax;
    pair<int , int> cs;
    if(collect[p.first][p.second] > 0)
        return collect[p.first][p.second];

    for(i = 0; i < 4; i++)
    {//对四个方向进行遍历
        cs = p;
        for(j = 0; j < k; j++)
        {
            cs.first += dir[i][0];
            cs.second += dir[i][1];
            if(!isBound(cs.first, cs.second))
                break;//越界，没有必要再继续走下去
            if(graph[cs.first][cs.second] > graph[p.first][p.second])
            {//此为满足条件2
                tmax = memory(cs);
                if(tmax > maxb)
                    maxb = tmax;
            }
        }
    }
    collect[p.first][p.second] = maxb + graph[p.first][p.second];
    return collect[p.first][p.second];
}
int main()
{
    pair<int, int> p;
    int i, j;
    while(scanf("%d%d", &n, &k))
    {
        if(n == -1 && k == -1)
            break;//本次输入结束
        for(i = 0; i < n; i++)
            for(j = 0; j < n; j++)
            {
                scanf("%d", &graph[i][j]);
                collect[i][j] = 0;
            }
        /**////////////////对零点进行初始化////////////////////
        p.first = p.second = 0;
        printf("%d\n", memory(p));
    }
    return 0;
}