c++ - 网格中的相邻产品

Question

我正在解决欧拉计划的第 11 题。我已经弄清楚算法以及我需要做什么。网格保存在文件 grid.txt 中，其内容是-

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

问题是 - 在 20x20 网格中任意方向(上、下、左、右或对角线)的四个相邻数字的最大乘积是多少？

我知道该算法工作正常，因为我尝试使用 cout 输出数字并且它们以正确的顺序显示。虽然它给了我一个错误的答案，可能是什么问题？

   void problem11()
{
    vector<vector<int>> grid;

    ifstream stream("grid.txt");
    string line;

    char *tok;

    if (stream.is_open())
    {
        while(stream.good())
        {
            getline(stream, line);

            tok = strtok((char *)line.c_str(), " ");

            vector<int> row;

            while (tok != NULL)
            {
                int field;
                stringstream ss;
                ss << tok;
                ss >> field;
                row.push_back(field);

                tok = strtok(NULL, " ");
            }
            grid.push_back(row);
        }
        stream.close();
    }

    unsigned long highest = 0;

    /// LEFT TO RIGHT
    for (int i=0; i < 20; i++) // i'th row
    {
        vector<int> row = grid.at(i);

        for (int c=0; c < 20-3; c++) // -3 to accomodate for last
        {
            unsigned long prod = row.at(c) * row.at(c+1) * row.at(c+2) * row.at(c+3); // four consecutive
            //cout << row.at(c) << " " << row.at(c+1) << " " << row.at(c+2) << " " << row.at(c+3) << endl;
            if (prod > highest)
                highest = prod;
        }
    }

    /// TOP TO DOWN
    /// This moves from left to right, then top to botom
    ///
    for (int i=0; i < 20-3; i++) // subtract last 3
    {
        vector<int> row1, row2, row3, row4;

        row1 = grid.at(i);
        row2 = grid.at(i+1);
        row3 = grid.at(i+2);
        row4 = grid.at(i+3);

        for (int c=0; c < 20; c++)
        {
            unsigned long prod = row1.at(c) * row2.at(c) * row3.at(c) * row4.at(c);
            //cout << row1.at(c) << " " << row2.at(c) << " " << row3.at(c) << " " << row4.at(c) << endl;
            if (prod > highest)
                highest = prod;
        }
    }

    /// DOWN DIAGONAL
    /// This moves diagonally from left to right, top to bottom
    for (int i=0; i < 20-3; i++) // subtract last 3
    {
        vector<int> row1, row2, row3, row4;

        row1 = grid.at(i);
        row2 = grid.at(i+1);
        row3 = grid.at(i+2);
        row4 = grid.at(i+3);

        for (int c=0; c < 20-3; c++) // omit last 3
        {
            unsigned long prod = row1.at(c) * row2.at(c+1) * row3.at(c+2) * row4.at(c+3);
            //cout << row1.at(c) << " " << row2.at(c+1) << " " << row3.at(c+2) << " " << row4.at(c+3) << endl;
            if (prod > highest)
                highest = prod;
        }
    }

    /// UP DIAGONAL
    /// This moves diagonally from left to right, bottom to top
    for (int i=3; i < 20; i++) // start from 3, skipping first four
    {
        vector<int> row1, row2, row3, row4;

        row4 = grid.at(i);
        row3 = grid.at(i-1);
        row2 = grid.at(i-2);
        row1 = grid.at(i-3);

        for (int c=0; c < 20-3; c++) // omit last 3
        {
            unsigned long prod = row4.at(c) * row3.at(c+1) * row3.at(c+2) * row4.at(c+3);
            //cout << row4.at(c) << " " << row3.at(c+1) << " " << row2.at(c+2) << " " << row1.at(c+3) << endl;
            if (prod > highest)
                highest = prod;
        }
    }

    cout << "Required: " << highest;

}

Answer 1

冒着破坏自己寻找答案的乐趣的风险......

打印出对角线。目视检查它们是否符合您的预期。

作为旁注:不要创建表格行的拷贝，但同样可以访问它们:vector[rowindex][column]。

编辑——

好的，现在我要破坏它了。在 NxN 的矩阵中，你有多少条对角线路径？你走过多少条路？ (通过采用 2x2 矩阵交叉检查，每个方向有 3 条对角线路径)

附言。如果您认真对待编程，遇到错误时，请先验证您的假设。