我试图遍历一个迷宫并产生所有有效路径。迷宫总是从S开始。所有出口都可以在最右边的任何地方 壁。顶壁和底壁总是实心的＃符号行。我只能穿过并且相邻的广场包含一个不是＆＃34;＃＆＃34;。
的字符这是一个迷宫的例子。
######
Sa#hln
#bdp##
##e#ko
#gfij#
######
这应该打印出来：
S，A，B，d，E，F，I，J，K，O
S，A，B，d，P，H，L，N
但我得到S，a，b，d，e，f，i，j，k，o，p，h，l，n
这是第一条路径，也是分叉的第二条路径的一半。我将不胜感激任何帮助。谢谢。
#include <fstream>
#include <string>
#include <iostream>
#include <stack>
using namespace std;
//std::ifstream file("maze.txt");
string str;
string file_contents;
fstream in("maze.txt");
//check if char is not #
bool characterAt(string maze[][6], int r, int c){
if (maze[r][c] == "#"){
return false;
}
return true;
}
//check if I have been there before
bool backTrack(bool maze[][6], int r, int c){
if (maze[r][c] == false){
return true;
}
return false;
}
//check if my move is in bounds.
bool validMove(int r, int c){
if (r - 1 < 0 || r + 1 > 6 ||
c - 1 < 0 || c + 1 > 6){
return false;
}
return true;
}`
//check if there is a next move
bool thereIsNextMove(string maze[][6], bool bmaze[][6], int rows, int columns){
if (validMove(rows - 1, columns) && backTrack(bmaze, rows - 1, columns) && characterAt(maze, rows - 1, columns) ||
validMove(rows + 1, columns) && backTrack(bmaze, rows + 1, columns) && characterAt(maze, rows + 1, columns) ||
validMove(rows, columns - 1) && backTrack(bmaze, rows, columns - 1) && characterAt(maze, rows, columns - 1) ||
validMove(rows, columns + 1) && backTrack(bmaze, rows, columns + 1) && characterAt(maze, rows, columns + 1)){
return true;
}
return false;
}
void generate_all_paths(string maze[][6], bool bmaze[][6], int rows, int columns){
int y = rows;
int x = columns;
string path = maze[y][x];
stack<string> s;
if (thereIsNextMove(maze, bmaze, y, x) || x == 5){
s.push(path);
cout << s.top();
s.pop();
}
if (validMove(y - 1, x) && characterAt(maze, y - 1, x) && backTrack(bmaze, y - 1, x)){
bmaze[y][x] = true;
generate_all_paths(maze, bmaze, y - 1, x);
}
if (validMove(y + 1, x) && characterAt(maze, y + 1, x) && backTrack(bmaze, y + 1, x)){
bmaze[y][x] = true;
generate_all_paths(maze, bmaze, y + 1, x);
}
if (validMove(y, x - 1) && characterAt(maze, y, x - 1) && backTrack(bmaze, y, x - 1)){
bmaze[y][x] = true;
generate_all_paths(maze, bmaze, y, x - 1);
}
if (validMove(y, x + 1) && characterAt(maze, y, x + 1) && backTrack(bmaze, y, x + 1)){
bmaze[y][x] = true;
generate_all_paths(maze, bmaze, y, x + 1);
}
}
int main(){
while (getline(in, str))
{
file_contents += str;
//file_contents.push_back('\n');
}
string maze[6][6];
for (int row = 0; row < 6; row++){
for (int col = 0; col < 6; col++){
char mazeChar = file_contents.at(6 * row + col);
maze[row][col] = mazeChar;
}
}
bool bmaze[6][6];
for (int brow = 0; brow < 6; brow++){
for (int bcol = 0; bcol < 6; bcol++){
bmaze[brow][bcol] = false;
}
}
string path;
generate_all_paths(maze, bmaze, 1, 0);
int age;
cin >> age;
return 0;
}
答案 0 :(得分：0)
这里的堆栈是＆＃39; generate_all_paths＆＃39;功能尚未使用。在给定的实现中，堆栈在任何给定时间最多包含一个元素。不是每次找到路径时都打印路径，只需保持堆栈全局并将新路径推送到它，然后在到达目标时立即打印整个堆栈，然后在函数末尾从堆栈中删除元素，就像你实际穿越迷宫一样，如何走回叉点。
#include <fstream>
#include <string>
#include <iostream>
#include <stack>
using namespace std;
//std::ifstream file("maze.txt");
string str;
string file_contents;
fstream in("maze.txt");
stack<string> s;
//check if char is not #
bool characterAt(string maze[][6], int r, int c){
if (maze[r][c] == "#"){
return false;
}
return true;
}
//check if I have been there before
bool backTrack(bool maze[][6], int r, int c){
if (maze[r][c] == false){
return true;
}
return false;
}
//check if my move is in bounds.
bool validMove(int r, int c){
if (r - 1 < 0 || r + 1 > 6 ||
c - 1 < 0 || c + 1 > 6){
return false;
}
return true;
}
//check if there is a next move
bool thereIsNextMove(string maze[][6], bool bmaze[][6], int rows, int columns){
if (validMove(rows - 1, columns) && backTrack(bmaze, rows - 1, columns) && characterAt(maze, rows - 1, columns) ||
validMove(rows + 1, columns) && backTrack(bmaze, rows + 1, columns) && characterAt(maze, rows + 1, columns) ||
validMove(rows, columns - 1) && backTrack(bmaze, rows, columns - 1) && characterAt(maze, rows, columns - 1) ||
validMove(rows, columns + 1) && backTrack(bmaze, rows, columns + 1) && characterAt(maze, rows, columns + 1)){
return true;
}
return false;
}
template < typename T > void print( const std::stack<T>& stk )
{
struct cheat : std::stack<T> { using std::stack<T>::c ; } ;
const auto& seq = static_cast< const cheat& >(stk).c ;
for( const auto& v : seq ) std::cout << v << ' ' ;
std::cout << '\n' ;
}
void generate_all_paths(string maze[][6], bool bmaze[][6], int rows, int columns){
int y = rows;
int x = columns;
string path = maze[y][x];
bool pushed = false;
if (thereIsNextMove(maze, bmaze, y, x)){
s.push(path);
pushed = true;
}
if (x == 5){
s.push(path);
pushed=true;
print(s);//printing the entire stack once a path has been found
}
if (validMove(y - 1, x) && characterAt(maze, y - 1, x) && backTrack(bmaze, y - 1, x)){
bmaze[y][x] = true;
generate_all_paths(maze, bmaze, y - 1, x);
}
if (validMove(y + 1, x) && characterAt(maze, y + 1, x) && backTrack(bmaze, y + 1, x)){
bmaze[y][x] = true;
generate_all_paths(maze, bmaze, y + 1, x);
}
if (validMove(y, x - 1) && characterAt(maze, y, x - 1) && backTrack(bmaze, y, x - 1)){
bmaze[y][x] = true;
generate_all_paths(maze, bmaze, y, x - 1);
}
if (validMove(y, x + 1) && characterAt(maze, y, x + 1) && backTrack(bmaze, y, x + 1)){
bmaze[y][x] = true;
generate_all_paths(maze, bmaze, y, x + 1);
}
if(pushed){
s.pop();//removing the element from the stack like how you walk back to fork point
}
}
int main(){
while (getline(in, str))
{
file_contents += str;
//file_contents.push_back('\n');
}
string maze[6][6];
for (int row = 0; row < 6; row++){
for (int col = 0; col < 6; col++){
char mazeChar = file_contents.at(6 * row + col);
maze[row][col] = mazeChar;
}
}
bool bmaze[6][6];
for (int brow = 0; brow < 6; brow++){
for (int bcol = 0; bcol < 6; bcol++){
bmaze[brow][bcol] = false;
}
}
string path;
generate_all_paths(maze, bmaze, 1, 0);
int age;
cin >> age;
return 0;
}
PS - 我添加了一个名为print的函数来打印堆栈