hdu1078 FatMouse and Cheese

题目链接

:http://acm.hdu.edu.cn/showproblem.php?pid=1078

题意:

老鼠每次最多只能走k步，停下的这个位置的数字比上一个停留的位置大，并获取其价值，每次只能水平或垂直走，问最大能得到的价值

简单DFS + 记忆化搜索
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#define maxn 105

using namespace std;

int n, k;
int ans;
int Map[maxn][maxn];
int dp[maxn][maxn];
int x[4] = {1, 0, -1, 0};
int y[4] = {0, 1, 0, -1};

int DFS(int r, int c)
{
    int Max = 0;
    //cout << dp[r][c] << endl;
    if(dp[r][c])
        return dp[r][c];
    for(int i = 1; i <= k; i++)
    {
        for(int j = 0; j < 4; j++)
        {
            int rr = r + i * x[j];
            int cc = c + i * y[j];
            if(rr >= n || rr < 0 || cc >= n || cc < 0)
                continue;
            if(Map[rr][cc] > Map[r][c])
            {
                Max = max(Max, DFS(rr, cc));
            }
        }
    }

    return  dp[r][c] = Max + Map[r][c];
}

int main()
{
    while(~scanf("%d %d", &n, &k))
    {
        if(n == -1 && k == -1)
            break;

        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                scanf("%d", &Map[i][j]);
            }
        }
        memset(dp, 0, sizeof(dp));
        ans = DFS(0, 0);
        cout << ans << endl;
    }
    return 0;
}