字解决-所有方向

这是一种有趣的思考方式:寻找单词类似于解决迷宫。'start'和'end'对应的是你要查找的单词的开头和结尾，'dead end'对应的是路径和你要查找的单词不匹配，'success'则表示路径上的字符串匹配。

好消息是有很多关于解迷宫算法的资源。有一种算法我很熟悉，而且实现起来并不难，那就是带回溯的递归。

显然，为了解决您的问题，必须进行一些更改。例如，你不知道起点在哪里，但幸运的是，这并不重要。您可以检查每一个可能的起始位置，并且由于不匹配，它们中的许多将在第一步中被丢弃。

1)目前，您的solve()函数在每个点开始的直线中查找单词:这是您想要的吗?我只是问，因为'hello'在您的示例矩阵中不显示为直线:

H  E  i  l
x  L  p  q
c  L  O  m

如果你只想要直线单词，那么很好(这就是我一直理解这些谜题的工作方式)，但如果实际上你想要以蛇形方式查找单词，那么像Zilchonum和BlueRaja建议的递归搜索将是一个不错的选择。只是要注意不要循环使用已经使用过的字母。

2)在任何一种情况下，你的verifyWord()函数也有一些问题:至少它需要在你退出while (low < high)循环的情况下返回一些值。

即便如此，它仍然不能完全做到你想要它做的事情:例如，你的字典包含{"ant", "bat" "hello", "yak", "zoo"}，并且您用str="hel"调用verifyWord()，您想返回值2，但目前它这样做:

step  low   mid  high
 0     0     0     5   // initialise
 1     0     2     5   // set mid = (0+5)/2 = 2... words[2] == "hello" 
 2     0     2     1   // "hel" < "hello" so set high = mid - 1
 3     0     0     1   // set mid = (0+1)/2 = 0... words[0] == "ant"
 4     1     0     1   // "hel" > "ant" so set low = mid + 1     
 5  // now (low<high) is false, so we exit the loop with mid==0

与其将"hel"与"hello"进行比较，不如将字典中的单词截断为与str相同的长度:即比较str与word[mid].substr(0,str.length()) ?

将其视为每个字母连接到所有相邻字母的图，并从每个字母开始进行深度/宽度优先搜索，只接受那些字母等于您要查找的下一个字母的节点

这是我编写的一个简单的单词侦查程序——>

#include<iostream>
using namespace std;
int main()
{
    int a, b, i, j, l, t, n, f, g, k;
    cout<<"Enter the number of rows and columns: "<<endl;               
    cin>>a>>b;                                                              //Inputs the number of rows and columns
    char mat[100][100], s[100];
    cout<<"Enter the matrix: "<<endl;
    for (i = 0; i < a; i++) for (j = 0; j < b; j++) cin>>mat[i][j];         //Inputs the matrix
    cout<<"Enter the number of words: "<<endl;
    cin>>t;                                                                 //Inputs the number of words to be found
    while (t--)
    {
        cout<<"Enter the length of the word: "<<endl;
        cin>>n;                                                             //Inputs the length of the word
        cout<<"Enter the word: "<<endl;
        for (i = 0; i < n; i++) cin>>s[i];                                  //Inputs the word to be found
        for (i = 0; i < a; i++)                                         //Loop to transverse along i'th row
        {
            for (j = 0; j < b; j++)                                     //Loop to transverse along j'th column
            {
                f = i;
                g = j;
                for (k = 0; s[k] == mat[f][g] && k < n; k++, g++);          //Loop to find the word if it is horizontally right
                if (k == n)
                {
                    cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" right"<<endl;
                    goto A;
                }
                f = i;
                g = j;
                for (k = 0; s[k] == mat[f][g] && k < n; k++, g--);      //Loop to find the word if it is horizontally left
                if (k == n)
                {
                    cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" left"<<endl;
                    goto A;
                }
                f = i;
                g = j;
                for (k = 0; s[k] == mat[f][g] && k < n; k++, f++);      //Loop to find the word if it is vertically down
                if (k == n)
                {
                    cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" down"<<endl;
                    goto A;
                }
                f = i;
                g = j;
                for (k = 0; s[k] == mat[f][g] && k < n; k++, f--);      //Loop to find the word if it is vertically up
                if (k == n)
                {
                    cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" up"<<endl;
                    goto A;
                }
                f = i;
                g = j;
                for (k = 0; s[k] == mat[f][g] && k < n; k++, f++, g++); //Loop to find the word if it is down right
                if (k == n)
                {
                    cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" down right"<<endl;
                    goto A;
                }
                f = i;
                g = j;
                for (k = 0; s[k] == mat[f][g] && k < n; k++, f--, g--); //Loop to find the word if it is up left
                if (k == n)
                {
                    cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" up left"<<endl;
                    goto A;
                }
                f = i;
                g = j;
                for (k = 0; s[k] == mat[f][g] && k < n; k++, f++, g--); //Loop to find the word if it is down left
                if (k == n)
                {
                    cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" down left"<<endl;
                    goto A;
                }
                f = i;
                g = j;
                for (k = 0; s[k] == mat[f][g] && k < n; k++, f--, g++); //Loop to find the word if it is up right
                if (k == n)
                {
                    cout<<"The coordinates and direction are ---> "<<j+1<<","<<i+1<<" up right"<<endl;
                    goto A;
                }
            }
        }
        A:;                                                             //If the word has been found the program should reach this point to start the search for the next word
    }
    return 0;
}

在我的程序中，它首先检查单词的第一个字母，然后检查后面的字母。如果找到该单词，则打印该单词的起始坐标和找到该单词的方向。