算法————回溯法--迷宫问题（残缺）

#include<iostream>
using namespace std;
int** path;
int** result;
int** square;
int dirx[8] = { 0,0,1,1,1,-1,-1,-1 };//8个方向变量 
int diry[8] = { 1,-1,-1,0,1,-1,0,1 };
int turn(999);
int level(0);
int finalLevel(0);

int* dir;//步向
int m, n;//迷宫大小
int p, q, r, s;//罗密哦朱丽叶坐标



bool trackback(int p, int q);
bool IsFull();
void main() {
	
	cout << "输入迷宫大小：" << endl;
	cin >> m >> n;
	
	cout << "输入罗密欧与朱丽叶坐标：" << endl;
	cin >> p >> q >> r >> s;

	//记录
	result = new int* [2];
	result[0]=new int[n * m];
	result[1] = new int[n * m];
	
	path = new int* [2];
	path[0] = new int[n * m];
	path[1] = new int[n * m];
	
	//记录每一步的方向
	
	dir = new int[m * n];

	//储存迷宫
	
	square = new int* [m + 2];
	for (int i = 0;i < m + 2;i++) {
		square[i] = new int[n + 2];
	}
	for (int i = 0;i < m + 2;i++) {
		for (int j = 0;j < n + 2;j++)
			square[i][j] = 0;
	}
	int k,x,y;
	cout << "输入封闭房间个数：" << endl;
	cin >> k;
	cout << "输入封闭房间的坐标;" << endl;
	for (int i = 0;i < k;i++) {
		cin >> x >> y;
		square[x][y] = -1;
	}
	for (int i = 0, j = 0;i < m + 2;i++) {
		square[i][j] = 1;
	}
	for (int i = 0, j = 0;j < n + 2;j++) {
		square[i][j] = 1;
	}
	for (int i = 0, j = n + 1;i<m+2;i++) {
		square[i][j] = 1;
	}
	for (int i = m + 1, j = 0;j < n + 2;j++) {
		square[i][j] = 1;
	}
	dir[0] = -1;
	trackback(p, q);
	cout << turn - 1 << endl;
	int i;
	for (i = 0;i < finalLevel;i++) {
		square[result[0][i]][result[1][i]] = i + 1;
	}
	for (i = 1;i < m + 1;i++) {
		for (int j = 1; j < n + 1;j++) {
			cout << square[i][j] << " ";
			
		}
		cout << endl;
	}
}
bool trackback(int p, int q) {

	path[0][level]=p;
	path[1][level] = q;
	level++;
	square[p][q] = 1;
	if (p==r&&q==s){
		if (IsFull()) {
			int count(0);
			for (int i = 1;i < level;i++) {
				if (dir[i] != dir[i - 1]) {
					count++;
				}
				if (count < turn) {
					finalLevel = level;
					turn = count;
					for (int i = 0;i < level;i++) {
						result[0][i] = path[0][i];
						result[1][i] = path[1][i];
					}
				}
			}
		}
		square[p][q] = 0;
		level--;
		return false;
	}
	for (int i = 0;i < 8;i++) {
		int x, y;
		x = p + dirx[i];
		y = q + diry[i];
		if (square[x][y]==0) {
			dir[level] = i;
			trackback(x, y);

		}
	}
	square[p][q] = 0;
	level--;
	return true;
}
bool IsFull() {
	for (int i = 1;i <= m;i++) {
		for (int j = 1;j <= n;j++) {
			if (square[i][j] == 0) {
				return false;
			}
			else {
				return true;
			}
		}

	}
}

　　trackback:

不完善，会出现无法便利的状况，我明明设计了isfull函数