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词搜索算法分割错误

[英]Word search algorithm segmentation fault

i've been working on word search algorithm for quite long I think i made it good and decided to test limits. 我一直在研究单词搜索算法很长时间,我认为我做得很好,并决定测试极限。 I've created program which makes file as big as I want to. 我已经创建了程序,该程序可以使文件尽可能的大。 So i made a matrix 10000 * 10000 (10000000 letters) and really long word from top left corner to bottom right corner. 所以我做了一个矩阵10000 * 10000(10000000个字母),并且从左上角到右下角都有很长的单词。 Thing is that it works with 4000 * 4000 matrix but then it gets bigger it just crashes. 事实是,它可以与4000 * 4000矩阵一起使用,但随后变得更大,它崩溃了。 I tried to comment all other checkings for possible location and left the right one and it works perfectly even with 10000 * 10000 matrix but as soon as I add other checks it stops and I have no idea why. 我试图注释所有其他检查以查找可能的位置,并在右边进行左对齐,即使使用10000 * 10000矩阵,它也可以正常工作,但是一旦添加其他检查,它就会停止并且不知道为什么。 Any suggestions? 有什么建议么?

My code: 我的代码:

    #include <iostream>     //Might Be:
    #include <string>       // <----->
    #include <fstream>      // /-\  (1)/\               /\(3)
    #include <new>          //  |       \               /
    #include <cstdlib>      //  |        \             /
                            //  |         \           /
                            //  |          \         /
                            //  |           \       /
                            // \_/       (2)\/     \/(4)
                            //

    using namespace std;
                                        //Loop[4] //Loop[5]
    int * Possibles(int Widht, int Height, int Poz, int Poz1, int Leng, int * Possible)
    {
        if(Poz1 < Widht - Leng + 1) // To right
        {
            Possible[0] = 1;
        }

        if(Poz1 >= Leng - 1) // To left
        {
            Possible[1] = 1;
        }

        if(Poz <= Height - Leng) // From top to bottom
        {
            Possible[2] = 1;
        }

        if(Poz >= Leng) // From bottom to top
        {
            Possible[3] = 1;
        }

        if(Poz + Leng <= Height && Poz1 + Leng <= Widht) //(2)
        {
            Possible[4] = 1;
        }

        if(Poz + Leng <= Height && Poz1 - Leng + 1 >= 0) //(4)
        {
            Possible[5] = 1;
        }

        if(Poz - Leng + 1 >= 0 && Poz1 - Leng + 1 >= 0) //(1)
        {
            Possible[6] = 1;
        }

        if(Poz - Leng + 1 >= 0 && Poz1 + Leng <= Widht) //(3)
        {
            Possible[7] = 1;
        }

        return Possible;
    }

    int * Zero(int * Possible)
    {
            Possible[0] = 0;
            Possible[1] = 0;
            Possible[2] = 0;
            Possible[3] = 0;
            Possible[4] = 0;
            Possible[5] = 0;
            Possible[6] = 0;
            Possible[7] = 0;

            return Possible;
    }

    string Next(string * NewMatrix, int Height, int Widht)
    {
        return NewMatrix[Height].substr(Widht, 1);
    }

    bool Find(string Word, int Poz, int Poz1, int Look, string Have, string * Matrix, int * Possible, int Backup, int Backup1)
    {
        if(Have == Word)
        {
            return true;
            return Possible;
        }

        string NewLet = Word.substr(Look, 1);

        if(Possible[0] == 1)
        {
            if(NewLet == Next(Matrix, Poz, Poz1 + 1))
            {
                Have += NewLet;

                return Find(Word, Poz, Poz1 + 1, Look + 1, Have, Matrix, Possible, Backup, Backup1);
            }
            else
            {
                Possible[0] = 0;
                Have = Word.substr(0, 1);

                return Find(Word, Backup, Backup1, 1, Have, Matrix, Possible, Backup, Backup1);
            }
        }

        if(Possible[1] == 1)
        {
            if(NewLet == Next(Matrix, Poz, Poz1 - 1))
            {
                Have += NewLet;

                return Find(Word, Poz, Poz1 - 1, Look + 1, Have, Matrix, Possible, Backup, Backup1);
            }
            else
            {
                Possible[1] = 0;
                Have = Word.substr(0, 1);

                return Find(Word, Backup, Backup1, 1, Have, Matrix, Possible, Backup, Backup1);
            }
        }

        if(Possible[2] == 1)
        {
            if(NewLet == Next(Matrix, Poz + 1, Poz1))
            {
                Have += NewLet;

                return Find(Word, Poz + 1, Poz1, Look + 1, Have, Matrix, Possible, Backup, Backup1);
            }
            else
            {
                Possible[2] = 0;
                Have = Word.substr(0, 1);

                return Find(Word, Backup, Backup1, 1, Have, Matrix, Possible, Backup, Backup1);
            }
        }

        if(Possible[3] == 1)
        {
            if(NewLet == Next(Matrix, Poz - 1, Poz1))
            {
                Have += NewLet;

                return Find(Word, Poz - 1, Poz1, Look + 1, Have, Matrix, Possible, Backup, Backup1);
            }
            else
            {
                Possible[3] = 0;
                Have = Word.substr(0, 1);

                return Find(Word, Backup, Backup1, 1, Have, Matrix, Possible, Backup, Backup1);
            }
        }

        if(Possible[4] == 1)
        {
            if(NewLet == Next(Matrix, Poz + 1, Poz1 + 1))
            {
                Have += NewLet;

                return Find(Word, Poz + 1, Poz1 + 1, Look + 1, Have, Matrix, Possible, Backup, Backup1);
            }
            else
            {
                Possible[4] = 0;
                Have = Word.substr(0, 1);

                return Find(Word, Backup, Backup1, 1, Have, Matrix, Possible, Backup, Backup1);
            }
        }

        if(Possible[5] == 1)
        {
            if(NewLet == Next(Matrix, Poz + 1, Poz1 - 1))
            {
                Have += NewLet;

                return Find(Word, Poz + 1, Poz1 - 1, Look + 1, Have, Matrix, Possible, Backup, Backup1);
            }
            else
            {
                Possible[5] = 0;

                Have = Word.substr(0, 1);

                return Find(Word, Backup, Backup1, 1, Have, Matrix, Possible, Backup, Backup1);
            }
        }

        if(Possible[6] == 1)
        {
            if(NewLet == Next(Matrix, Poz - 1, Poz1 - 1))
            {
                Have += NewLet;

                return Find(Word, Poz - 1, Poz1 - 1, Look + 1, Have, Matrix, Possible, Backup, Backup1);
            }
            else
            {
                Possible[6] = 0;
                Have = Word.substr(0, 1);

                return Find(Word, Backup, Backup1, 1, Have, Matrix, Possible, Backup, Backup1);
            }
        }

        if(Possible[7] == 1)
        {
            if(NewLet == Next(Matrix, Poz - 1, Poz1 + 1))
            {
                Have += NewLet;

                return Find(Word, Poz - 1, Poz1 + 1, Look + 1, Have, Matrix, Possible, Backup, Backup1);
            }
            else
            {
                Possible[7] = 0;
                Have = Word.substr(0, 1);

                return Find(Word, Backup, Backup1, 1, Have, Matrix, Possible, Backup, Backup1);
            }
        }

        return false;
    }

    string Diro(int * Possible)
    {
        string Dir;

        bool Next = true;

        if(Possible[0] == 1 && Next == true)
        {
            Dir = " From right to left";
            Next = false;
        }

        if(Possible[1] == 1 && Next == true)
        {
            Dir = " From left to right";
            Next = false;
        }

        if(Possible[2] == 1 && Next == true)
        {
            Dir = " From top to bottom";
            Next = false;
        }

        if(Possible[3] == 1 && Next == true)
        {
            Dir = " From bottom to top";
            Next = false;
        }

        if(Possible[4] == 1 && Next == true)
        {
            Dir = " ";
            Next = false;
        }

        if(Possible[5] == 1 && Next == true)
        {
            Dir = " ";
            Next = false;
        }

        if(Possible[6] == 1 && Next == true)
        {
            Dir = " ";
            Next = false;
        }

        if(Possible[7] == 1 && Next == true)
        {
            Dir = " ";
            Next = false;
        }

        return Dir;
    }

    int main()
    {
        int Height = 0, Widht = 0, Numb = 0;

        int Loop[] = {0, 0, 0, 0, 0, 0, 0, 0, 0};

        int * Possible = new int[8];

        string Dir, Search, Tempo, Temp;

        ifstream Data("C:/Users/Magician/AppData/Local/VirtualStore/Program Files (x86)/CodeBlocks/MakeMaze/Files/Maze.txt");

        Data >> Widht >> Height;

        string * NewMatrix = new string[Height];

        while(Loop[7] < Height)
        {
            Tempo = "";
            Loop[8] = 0;

            while(Loop[8] < Widht)
            {
                Data >> Temp;
                Tempo += Temp;
                Loop[8]++;
            }

            NewMatrix[Loop[7]] = Tempo;

            Loop[7]++;
        }

        Data >> Numb;

        string * Words = new string[Numb];

        while(Loop[2] < Numb)
        {
            Data >> Words[Loop[2]];

            Loop[2]++;
        }

        Data.close();

        while(Loop[3] < Numb)
        {
            Search = Words[Loop[3]].substr(0, 1);
            Loop[4] = 0;

            while(Loop[4] < Height)
            {
                Loop[5] = 0;

                while(Loop[5] < Widht)
                {
                    if(NewMatrix[Loop[4]].substr(Loop[5], 1) == Search)
                    {
                        Zero(Possible);
                        Possibles(Widht, Height, Loop[4], Loop[5], Words[Loop[3]].size(), Possible);

                        if(Find(Words[Loop[3]], Loop[4], Loop[5], 1, Search, NewMatrix, Possible, Loop[4], Loop[5]))
                        {
                            cout << Words[Loop[3]] << " At: " << Loop[4] + 1 << " collumn, symbol " << Loop[5] + 1 << " " << Diro(Possible) << endl;

                            Loop[5] = Widht;
                            Loop[4] = Height;
                        }
                    }

                    Loop[5]++;
                }

                Loop[4]++;
            }

            Loop[3]++;
        }

        delete [] Possible;
        delete [] Words;
        delete [] NewMatrix;

        return 0;
    }

If you didn't understood what I wrote before: when I comment every if(Possible[] == ) except for if(Possible[5] == 1) in function Find() algorithm works then all allowed it doesn't. 如果您不了解我之前写的内容:当我对函数if(Possible [5] == 1)中的所有if(Possible [] == 1)进行注释时,所有允许的注释都无效。 I've tried with 100 * 100 matrix with a lot of words to find and everything's ok. 我已经尝试使用100 * 100矩阵来查找很多单词,并且一切正常。

  1. One condition in Possibles is incorrect: Possibles一个条件不正确:

     /* INCORRECT: Should be [ Poz >= Leng - 1 ] */ if(Poz >= Leng) // From bottom to top { Possible[3] = 1; } 

    But this one is only a logical error and should not cause segmentation faults. 但是,这只是一个逻辑错误,不应引起分段错误。

  2. It looks like you have encountered a stack overflow. 看来您遇到了堆栈溢出。

    Let's do simple calculation. 让我们做简单的计算。 For 10000 * 10000 matrix and word length of 10000, if you start calling Find() at the top left of the matrix, then three directions are possible. 对于10000 * 10000矩阵和10000字长,如果您开始在矩阵的左上方调用Find() ,则可能会出现三个方向。 In worst case, Find() will traverse about 10000*3 elements. 在最坏的情况下, Find()将遍历大约10000 * 3个元素。 Note in Func() there are 3 string instances (sizeof(string) == 24 in 32bit VC2013), plus various integers. 请注意,在Func()有3个字符串实例(在32位VC2013中为sizeof(string)== 24),以及各种整数。 The size of a single frame could easily exceed 100 bytes. 单帧的大小很容易超过100个字节。 Since you are using recursive calls, this could lead to a stack usage of at least 10000 * 3 * 100 = 3000000bytes = approx. 由于您正在使用递归调用,因此这可能导致至少10000 * 3 * 100 = 3000000bytes =大约使用堆栈。 3M. 3M。

    This number is not very large, but enough for a stack overflow since Windows has a default stack size of 1M. 这个数字不是很大,但是足以使堆栈溢出,因为Windows的默认堆栈大小为1M。 ( http://msdn.microsoft.com/en-us/library/8cxs58a6.aspx ) http://msdn.microsoft.com/en-us/library/8cxs58a6.aspx

Advice for improvements 改进建议

This is my used pattern to solve this kinds of matrix traversal problems. 这是我用来解决此类matrix traversal问题的模式。

First, define an constant array to hold offsets for movements ( Moore neighborhood ): 首先,定义一个常数数组来保存运动的偏移量( 摩尔邻域 ):

const int delta[8][2] = {
    { 1, 0 }, { 1, 1 }, { 0, 1 }, { -1, 1 },
    { -1, 0 }, { -1, -1 }, { 0, -1 }, { 1, -1 }
};

Second, use a single for to check for all directions: 其次,使用单一for检查各个方向:

int initial_x = .., initial_y = ..;
for (int dir = 0; dir < 8; dir++) {
    for (int count = 0; count < WORD_LENGTH; count++) {
        int current_x = initial_x + delta[dir][0] * count;
        int current_y = initial_y + delta[dir][1] * count;
        if (IS_INVALID(current_x, current_y)) {
            break;
        }
    }
}

Last, insert various code and flags to complete the program. 最后,插入各种代码和标志以完成程序。

Another hint: you can use char type to get and compare a single character in a string (Use word[idx] to get idx th character of word ). 另一个提示:您可以使用char类型来获取和比较字符串中的单个字符(使用word[idx]获得idxword个字符)。 This could be substantially faster than using substr . 这可能比使用substr快得多。

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