迷宫问题（bfs+路径记录）

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_39562952/article/details/82701833

思路：正常bfs搜索，然后记录每个点的父节点，最后找父节点，最后输出路径；

代码如下：

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <cstring>
#include <climits>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stdio.h>
#define inf 429496729
using namespace std;
int then[4][2] = { {-1,0},{0,-1},{0,1},{1,0} };
struct fnode
{
	int x;
	int y;
}ans[100];
struct node
{
	int val;
	fnode ff;
	int x;
	int y;
	int step;
}ma[5][5];
bool maa[5][5];//记录有没有走过
int main()
{
	for (int i = 0; i < 5; i++)
	{
		for (int j = 0; j < 5; j++)
		{
			cin >> ma[i][j].val;
			ma[i][j].x = i;
			ma[i][j].y = j;
			
		}
	}
	queue<node>q;
	ma[0][0].step = 0;
	maa[0][0] = true;
	q.push(ma[0][0]);
	int flag = 0;
	while (!flag)
	{
		for (int i = 0; i < 4; i++)
		{
			int xx, yy;
			xx = q.front().x + then[i][0];
			yy = q.front().y + then[i][1];
			if (xx >= 5 || yy >= 5 || xx < 0 || yy < 0 || ma[xx][yy].val == 1 || maa[xx][yy]== true)
				continue;
			ma[xx][yy].step = ma[q.front().x][q.front().y].step+1;
			maa[xx][yy] = true;
			q.push(ma[xx][yy]);
			ma[xx][yy].ff.x = q.front().x;
			ma[xx][yy].ff.y = q.front().y;
			if (xx == 4 && yy == 4)
			{
				flag = 1;
				int ans_count = 0;
				fnode t;
				t.x = xx;
				t.y = yy;
				while (1)
				{
					ans[ans_count].x = t.x;
					ans[ans_count].y = t.y;
					ans_count++;
					if (t.x == 0 && t.y == 0)
						break;
					t = ma[t.x][t.y].ff;
					
				}
				for (int i = ans_count - 1; i >= 0; i--)
				{
					printf("(%d, %d)\n", ans[i].x, ans[i].y);
				}

			}
		}
		q.pop();
		
	}

	system("pause");

	return 0;
}