Question

我的C ++课程作业是对哈利波特的Tri Wizard Tournament迷宫挑战的解释。作业说明如下：http://www.cs.rhodes.edu/~kirlinp/courses/cs2/s17/proj/proj5/

目前我正在努力解决我的求解功能。移动范围超出范围，例如：大于行数/列数或负数的数字。我也得到无限的输出，就像动作一直持续直到被终止，因为被识别为无限循环。

如果您愿意查看我的代码/作业，我将非常感激。我觉得我的大部分代码都是正确的，但我有些问题，我不知道如何解决自己。

#include <iostream>
#include <vector>
#include <fstream>


using namespace std;

vector<string> read_function(string file);
string solve(vector<string> maze, int &row, int &col, int &numcalls);
void find_start(vector<string> maze, int &row, int &col);
void print_state();
vector<string> maze;
string s;

int main() {
    cout << "Enter a file name: " << endl;
    string file;
    cin >> file;

    vector<string> test = read_function(file);

    print_state();

    int row = 0;
    int col = 0;
    int numcalls = 0;

    find_start(maze, row, col);
    cout << "start is: " << row << "," << col << endl;

    string solution = solve(maze, row, col, numcalls);
    cout << solution << endl;
    cout << "Number of steps: " << numcalls << endl;

    print_state();

    return 0;
}

vector<string> read_function(string file) {
    ifstream infile; // Input file stream
    infile.open(file);

    while (getline(infile, s)) //
    {
        //infile >> s;
        maze.push_back(s);
    }
    infile.close();
    return maze;
}


void print_state(){
    for (int i = 0; i < maze.size(); i++){
        cout << maze[i] << endl;
    }
}

string solve(vector<string> maze, int &row, int &col, int &numcalls){
    numcalls += 1;
    cout << row << "," << col << endl;
    maze[row][col] = 'o';
    if (maze[row][col] == 'C') {
        return " C ";
    }
    if (row>=0 && maze[row-1][col] != '#' && maze[row-1][col] != 'o'&& maze[row-1][col] != ' '){    // MOVE NORTH Y/N
        row --;
        string answer = solve(maze, row, col, numcalls);
        if (answer != "X"){
            return "N"+answer;
        }
    }
    if (maze[row+1][col] != '#' && maze[row+1][col] != 'o' && maze[row+1]
[col] != ' '){   // MOVE SOUTH Y/N
        row ++;
        string answer = solve(maze, row, col, numcalls);
        if (answer != "X"){
             return "S"+answer;
        }
    }
    if (maze[row][col+1] != '#' && maze[row][col+1] != 'o' && maze[row]
[col+1] != ' '){   // MOVE EAST Y/N
        col ++;
        string answer = solve(maze, row, col, numcalls);
        if (answer != "X"){
            return "E"+answer;
        }
    }
    if (maze[row][col-1] != '#' && maze[row][col-1] != 'o' && maze[row][col-
1] != ' '){        // MOVE WEST Y/N
        col --;
        string answer = solve(maze, row, col, numcalls);
        if (answer != "X"){
            return "W"+answer;
        }
    }
    else{
        maze[row][col] = '.';
    }
    return "X";
}**

void find_start(vector<string> maze, int &row, int &col){

   for (int i = 0; i < maze.size(); i++) {
       for (int j = 0; j < maze[i].size(); j++){
           if (maze[i][j] == 'H'){
               row = i;
               col = j;
           }
        }

   }
}

Answer 1

缺少多个绑定检查，以及路径的返回字符串值将始终返回＆＃39; X＆＃39;。

绑定检查 - 如果访问数组，请确保您具有有效索引 - 在您的情况下，在数组访问中col = -1 / row + -1的每个条件都需要先检查（请参阅我的源示例）。这几乎是你要求的
你的面包屑标记覆盖了你的＆＃39; C＆＃39;标记。
按值将传递或行/列索引更改为递归方法调用 - 更容易跟踪其值
any_direction bool变量解析wheter转储完整路径（最终方法调用）或只是完成＆＃39; X＆＃39;

您的fixed（）方法的固定来源：

    string solve(vector<string>& maze, int row, int col, int &numcalls) {

    numcalls += 1;
    cout << row << "," << col << endl;
    string answer;

    bool any_direction = false;

    if (maze[row][col] == 'C') {
        return " C ";
    }
    maze[row][col] = 'o';

    if (row > 0 && maze[row-1][col] != '#' && maze[row-1][col] != 'o'&& maze[row-1][col] != ' '){    // MOVE NORTH Y/N
        //row--;
        answer = solve(maze, row-1, col, numcalls);
        if (answer != "X"){
            any_direction = true;
            return "N"+answer;
        }
    }
    if (row < (maze.size()-1) && maze[row+1][col] != '#' && maze[row+1][col] != 'o' && maze[row+1][col] != ' '){   // MOVE SOUTH Y/N
        //row++;
        answer = solve(maze, row+1, col, numcalls);
        if (answer != "X"){
            any_direction = true;
            return "S"+answer;
        }

    }
    if (col < (maze[row].size()-1) && maze[row][col+1] != '#' && maze[row][col+1] != 'o' && maze[row][col+1] != ' '){   // MOVE EAST Y/N
        //col++;
        answer = solve(maze, row, col+1, numcalls);
        if (answer != "X"){
            any_direction = true;
            return "E"+answer;
        }
    }
    if (col > 0 && maze[row][col-1] != '#' && maze[row][col-1] != 'o' && maze[row][col-1] != ' '){        // MOVE WEST Y/N
        //col--;
        answer = solve(maze, row, col-1, numcalls);
        if (answer != "X"){
            any_direction = true;
            return "W"+answer;
        }
    }

    if(!any_direction) {
        maze[row][col] = '.';
        return "X";
    }
    return answer;
}

此解决方案不适合您的任务，但希望它能帮助您完成它并更好地理解递归。享受您的C ++学习！

我的解决方法有什么问题？＃34;我的C ++作业中的函数？

1 个答案: