MatrixCode – Moonstone


MatrixCode – Moonstone
http://www.1point3acres.com/bbs/thread-189039-1-1.html
3 写一个function, 输入一个char[][], 一个int len。char[][] 里每个字符都是distinct的。输出int count。
count 的含义是 char array 里能生成的所有长度为len的code的个数。
code 的含义是可以通过char[][] 生成的字符串。code里的下一个字符可以是当前字符的上/下/左/右字符,也可以是它自己。但是不能是'*'或者'#'。
举个例子:
input:
char[][]:
a b c
d e f
* # x
len: 2
可能的code有:
aa, ab, ad, bb, ba, bc, be, cc, cb, cf, dd, da, de, ee, eb, ed, ef, ff, fc, fe, fx, xx
一共22个可能的code,
输出 count = 22。


一开始的解法穷举所有可能性。后来发现用dp可以做。
建一个memo
如果长度是一,所有位置都是一。
如果长度是二,每一个位置都是上一步中 自己+上+下+左+右。
。。。。
是因为每一步时每个位置只纪录从自己开始的。
memo + DP, 例如:. 鐣欏鐢宠璁哄潧-涓€浜╀笁鍒嗗湴
a, b, 
c, d
先考虑长度是1,那只能是每个字母自己,
1, 1
1, 1
长度是2的时候,对a来说可以是a + (a, b, c), 是3
3,3
3,3
长度是3的时候,对a来说可以是a + (aa, ab, ac), a + (b 和 c 长度是2 的变化)
9,9
9,9

public List<string> MatrixCode(char[,] matrix, int len)
{
    List<string> result = new List<string>();
    if ((matrix == null) || (len <= 0))
    {
        return result;
    }
    int R = matrix.GetLength(0);
    int C = matrix.GetLength(1);
    Dictionary<Tuple<int, int>, List<string>> dict = new Dictionary<Tuple<int, int>, List<string>>();
    for (int r = 0; r < R; r++)
    {
        for (int c = 0; c < C; c++)
        {
            if((matrix[r, c] != '*') && (matrix[r, c] != '#'))
            {
                int index = r * R + c;
                Tuple<int, int> key = new Tuple<int, int>(index, 1);
                dict.Add(key, new List<string>());
                dict[key].Add(matrix[r, c].ToString());
            }
        }
    }
    int[] dirX = new int[] { 0, 0, -1, 0, 1 };
    int[] dirY = new int[] { 0, -1, 0, 1, 0 };
    for (int k = 2; k <= len; k++)
    {
        for (int r = 0; r < R; r++)
        {
            for (int c = 0; c < C; c++)
            {
                int index = r * R + c;
                if ((matrix[r, c] != '*') && (matrix[r, c] != '#'))
                {
                    Tuple<int, int> key = new Tuple<int, int>(index, len);
                    if (!dict.ContainsKey(key))
                    {
                        dict.Add(key, new List<string>());
                    }
                    for (int i = 0; i <= 4; i++)
                    {
                        int m = r + dirX[i];
                        int n = c + dirY[i];
                        if ((m >= 0) && (m < R) && (n >= 0) && (n < C) &&
                            (matrix[m, n] != '*') && (matrix[m, n] != '#'))
                        {
                            Tuple<int, int> subKey = new Tuple<int, int>(m * R + n, k - 1);
                            if (dict.ContainsKey(subKey))
                            {
                                List<string> subValue = dict[subKey];
                                foreach (var s in subValue)
                                {
                                    string str = matrix[r, c] + s;
                                    if (!dict[key].Contains(str))
                                    {
                                        dict[key].Add(str);
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    for (int r = 0; r < R; r++)
    {
        for (int c = 0; c < C; c++)
        {
            Tuple<int, int> key = new Tuple<int,int>(r * R + c, len);
            if (dict.ContainsKey(key))
            {
                result.AddRange(dict[key]);
            }
        }
    }
    return result;
}

另问第三题如何用dp做,只写了dfs的
void codeString(set<string> &s, vector<vector<char>> vv, int len, int x, int y, string curStr). 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
{
        if (x < 0 || y < 0 || x >= (int)vv.size() || y >= (int)vv[0].size() || curStr.size() > len || vv[x][y] == '#' || vv[x][y] == '*')return;
        if (s.find(curStr) == s.end() && (int)curStr.size() == len)
        {
                s.insert(curStr);
                return;
        }
        codeString(s, vv, len, x, y, curStr + vv[x][y]);
        codeString(s, vv, len, x + 1, y,  curStr + vv[x][y]);
        codeString(s, vv, len, x, y + 1,  curStr + vv[x][y]);
        codeString(s, vv, len, x - 1, y,  curStr + vv[x][y]);
        codeString(s, vv, len, x, y - 1,  curStr + vv[x][y]);. 1point3acres.com/bbs
}
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