单词求解器 - 全方位
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【中文标题】单词求解器 - 全方位【英文标题】：Word Solver - All Directions 【发布时间】：2011-04-30 21:54:01 【问题描述】：我为所有方向创建了一个单词求解器。它可以水平、垂直和反向查找单词。但是，我在让它向各个方向发展时遇到了问题。所以带上“你好”：
H E i l
x L p q
c L O m
任何人都可以指点我如何做到这一点？这是我搜索单词的算法（在 C++ 中）：
/*
* For loops that search each row, each column in all 8 possible directions.
*/
void Scramble::solve()
cout << "Output:" << endl;
for (int row = 0; row < getRows(); row++)
for (int col = 0; col < getCols(); col++)
for (int rowDir = -1; rowDir <= 1; rowDir++)
for (int colDir = -1; colDir <=1; colDir++)
if (rowDir != 0 || colDir != 0)
findWords(row, col, rowDir, colDir);
/*
* Finds the matches in a given direction. Also calls verifyWord() to verify that the
* current sequence of letters could possibly form a word. If not, search stops.
*/
void Scramble::findWords(int startingRow, int startingCol, int rowDir, int colDir)
int searchResult;
string sequence = "";
sequence = sequence + wordsArr[startingRow][startingCol];
for (int i = startingRow + rowDir, j = startingCol + colDir; i >= 0 && j >= 0
&& i < getRows() && j < getCols(); i = i + rowDir, j = j + colDir)
sequence = sequence + wordsArr[i][j];
if (sequence.length() >= 3)
searchResult = verifyWord(words, sequence);
if ((unsigned int)searchResult == words.size())
break;
if (words[searchResult].rfind(sequence) > words[searchResult].length())
break;
if (words[searchResult] == (sequence))
cout << sequence << endl;
/*
* Performs the verifyWord search method.
* Searches the word to make sure that so far, there is possibly that the current sequence
* of letter could form a word. That is to avoid continuing to search for a word
* when the first sequence of characters do not construct a valid word in the dictionary.
*
* For example, if we have 'xzt', when this search is done it prevents the search
* to continue since no word in the dictionary starts with 'xzt'
*/
int Scramble::verifyWord(vector<string> words, string str)
int low = 0;
int mid = 0;
int high = words.size();
while (low < high)
mid = (low + high) / 2;
if (str > words[mid])
low = mid + 1;
else if (str < words[mid])
high = mid - 1;
else
return mid;
【问题讨论】：
【参考方案1】：这是一种有趣的思考方式：找到单词类似于解迷宫。 'start' 和 'end' 对应于您要查找的单词的开头和结尾，'dead end' 对应于路径与您的单词之间的不匹配，而 'success' 是沿着您的路径的字符串是匹配的。
这里的好消息是有很多关于解迷宫算法的资源。我熟悉且不难实现的一种特定算法是recursion with backtracking。
显然，必须进行一些更改才能解决您的问题。例如，您不知道起点在哪里，但幸运的是，这并不重要。你可以检查每一个可能的起始位置，其中许多会因为不匹配而在第一步中被丢弃。
【讨论】：
你可以说这是一个寻路问题。 :)【参考方案2】：1) 目前，您的solve() 函数从每个点开始查找直线的单词：这是您想要的吗？我只是问，因为“你好”在您的示例矩阵中没有显示为直线：
H E i l
x L p q
c L O m
如果你只想要 straight-line 单词，那很好（这就是我一直理解 these puzzles 无论如何工作的方式），但如果实际上你想在snake-wise 时尚，然后像 Zilchonum 和 BlueRaja 这样的递归搜索建议将是一个不错的选择。请注意，不要最终循环回您已经使用过的字母。
2) 无论哪种情况，您的verifyWord() 函数也存在一些问题：至少它需要在您退出while (low < high) 循环的情况下返回一些值。
即便如此，它仍然不能完全按照您的意愿行事：例如，说您的字典
包含"ant", "bat" "hello", "yak", "zoo"，而你用str="hel" 调用verifyWord()，你想返回一个值2，但目前它是这样做的：
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和word[mid].substr(0,str.length())？
【讨论】：
听起来不错 - 但它确实适用于直线。 是的，它仍然可以工作，但我认为findWords() 会尝试匹配非字典单词......它只会在到达矩阵边缘时停止尝试。 verifyWord() 将成功地将“hello”匹配到“hello”，所以当你最终找到这个词时，你会得到正确的结果。它只是做了很多额外的工作才能到达那里，因为您没有“避免在第一个字符序列没有在字典中构成有效单词时继续搜索单词。” 【参考方案3】：
只需将其视为一个图，其中每个字母都连接到所有相邻字母，并从每个字母开始进行深度/广度优先搜索，只接受那些字母等于您要查找的下一个字母的节点.
【讨论】：
你认为我可以通过在上面的代码中再添加一个递归函数来实现吗？【参考方案4】：这是我写的一个简单的单词侦探程序--->
#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，然后是后面的字母。如果找到单词，则打印单词的起始坐标和找到单词的方向。
【讨论】：
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