模拟求解迷宫问题（DFS+BFS）

2019年3月23日 317次阅读来源: N3verL4nd

代码下载：http://download.csdn.net/detail/lgh1992314/5326941

迷宫是实验心理学中一个古典问题。以一个n*m的长方阵表示迷宫，0和1分别表示迷宫中的通路和障碍。入口在左上方（1，1）处，出口在右下方（n，m）处。

要求求出从迷宫入口到出口有无通路的最短路径。

生成迷宫（调用的c语言中的srand和rand函数）：

#include <iostream>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <windows.h>
using namespace std;
#define MAX 1000
int map[MAX][MAX];

void CreateMap(int n, int m)              //1代表墙壁 0代表通路
{
      for(int i = 1; i <= n; i++)
      {
          for(int j = 1; j <= m; j++)
          {
              map[i][j] = rand() % 2;
              Sleep(5);
          }
      }
}

int main()
{
    freopen("map.txt", "w", stdout);
    int n, m;
    srand((unsigned)time(NULL));
    memset(map, -1, sizeof(map));
    cin >> n >> m;
    CreateMap(n, m);
    map[1][1] = 0;
    map[n][m] = 0;
    for(int i = 1; i <= n; i++)
      {
          for(int j = 1; j <= m; j++)
          {
              if(j != m)
              {
                   cout << map[i][j] << " ";
              }
              else
              {
                  cout << map[i][j] << endl;
              }
              
          }
      }
    return 0;
}

迷宫搜索：(BFS)

路径标记使用了API：SetConsoleTextAttribute

#include <iostream>
#include <queue>
#include <cstring>
#include <fstream>
#include <cstdio>
#include <Windows.h>
using namespace std;
#define MAX 1000
#define turn 8
typedef struct Point
{
	int x;
	int y;
	int cnt;
}Point;
const int moves[turn][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
int map[MAX][MAX];
bool visited[MAX][MAX];
int color[MAX][MAX];
int n, m, ans;
queue<Point>Q;
Point back[MAX][MAX], path[MAX];

void LoadMap()
{
	cin >> n >> m;
	fstream cin("map.txt");
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= m; j++)
		{
			cin >> map[i][j];
		}
	}
	cin.close();
}

void ShowMap()
{
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= m; j++)
		{
			if(j != m)
			{
				cout << map[i][j] << " ";
			}
			else
			{
				cout << map[i][j] << endl;
			}

		}
	}
}

void bfs(int x, int y)
{
	while(!Q.empty())
	{
		Q.pop();
	}
	Point pre, next;
	pre.x = x;
	pre.y = y;
	pre.cnt = 0;
	visited[pre.x][pre.y] = true;
	Q.push(pre);
	while(!Q.empty())
	{
		pre = Q.front();
		Q.pop();
		if(pre.x == n && pre.y == m)
		{
			ans = pre.cnt;
			break;
		}
		for(int i = 0; i < turn; i++)
		{
			next.x = pre.x + moves[i][0];
			next.y = pre.y + moves[i][1];
			if(next.x >= 1 && next.x <= n && next.y >= 1 && next.y <= m && !visited[next.x][next.y] && map[next.x][next.y] == 0)
			{
				back[next.x][next.y] = pre;
				next.cnt = pre.cnt + 1;
				visited[next.x][next.y] = true;
				Q.push(next);
			}
		}
	}
	for(int i = ans; i >= 0; i--)
	{
		path[i] = pre;
		pre = back[pre.x][pre.y];
	}
}

void ShowSetp()
{
	HANDLE handle;
	handle = GetStdHandle(STD_OUTPUT_HANDLE);
	memset(color, 0, sizeof(color));
	for(int i = 0; i <= ans; i++)
	{
		printf("(%d, %d)\n",path[i].x,path[i].y);
		color[path[i].x][path[i].y] = 1;
	}
	int count = 0;
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= m; j++)
		{
			if(color[i][j] == 1)
			{
				SetConsoleTextAttribute(handle,  
										FOREGROUND_GREEN |  
										FOREGROUND_INTENSITY);  
				cout << map[i][j] << " ";
			}
			else
			{
				SetConsoleTextAttribute(handle,  
					FOREGROUND_RED |  
					FOREGROUND_INTENSITY);  
				cout << map[i][j] << " ";
			}
		}
		cout << endl;
		count++;
	}
}

/*
void BcakTrack(int x, int y)
{
	if(back[x][y].x == 1 && back[x][y].y == 1)
	{
		printf("(1, 1)\n");
		return;
	}
	else
	{
		BcakTrack(back[x][y].x, back[x][y].y);
		printf("(%d, %d)\n", back[x][y].x, back[x][y].y);
	}
}*/																		

int main()
{
	LoadMap();
	ShowMap();
	memset(visited, false, sizeof(visited));
	ans = 0;
	bfs(1, 1);
	if(ans != 0)
	{
		cout << "It will take " << ans << " setps to the destination" << endl;
		ShowSetp();
		//BcakTrack(n, m);
	}
	else
	{
		cout << "Can't Find a way to the destination" << endl;
	}
	return 0;
}

测试程序(调用了bat，拒绝手动，主要还是懒=。=)

Monitoring.bat

@shift
@echo off
setlocal
MODE con: COLS=40 lines=15
color 0a
title  迷宫    by_N3verL4nd
:Start
for /L %%a in (
15,-1,0
) do (
echo.
echo.
echo.
echo.
echo    ☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆
echo    ☆  每隔15秒自动启动迷宫程序    ☆
echo    ☆        还剩余 %%a 秒           ☆
echo    ☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆
ping 127.0.0.1 -n 2 >nul 
cls
)
echo 23 39 | CreateMap.exe
ping 127.0.0.1 -n 2 >nul
start maze_bfs.bat
start maze_dfs.bat
goto Start

maze_bfs.bat

@echo off
title Maze_bfs...
echo 23 39 | maze_bfs.exe
ping 127.0.0.1 -n 10 > nul
exit

maze_dfs.bat

@echo off
title Maze_dfs...
echo 23 39 | maze_dfs.exe
ping 127.0.0.1 -n 10 > nul
exit

DFS：

#include <iostream>
#include <queue>
#include <cstring>
#include <fstream>
#include <cstdio>
#include <Windows.h>
using namespace std;
#define MAX 1000
#define turn 8
typedef struct Point
{
	int x;
	int y;
}Point;
const int moves[turn][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
int map[MAX][MAX];
bool visited[MAX][MAX];
int color[MAX][MAX];
int n, m, ans;
bool flag;
Point back[MAX][MAX], path[MAX];

void LoadMap()
{
	cin >> n >> m;
	fstream cin("map.txt");
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= m; j++)
		{
			cin >> map[i][j];
		}
	}
	cin.close();
}

void ShowMap()
{
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= m; j++)
		{
			if(j != m)
			{
				cout << map[i][j] << " ";
			}
			else
			{
				cout << map[i][j] << endl;
			}

		}
	}
}

void dfs(int x, int y)
{
	if(x == n && y == m)
	{
		flag = true;
		Point pre;
		pre.x = n;
		pre.y = m;
		while (back[pre.x][pre.y].x != 0 && back[pre.x][pre.y].y != 0)
		{
			pre = back[pre.x][pre.y];
			ans++;
		}
		//cout << "ans = " << ans <<  endl;
		pre.x = n;
		pre.y = m;
		for(int i = ans; i >= 0; i--)
		{
			path[i] = pre;
			pre = back[pre.x][pre.y];
		}
		for(int i = 0; i <= ans; i++)
		{
			printf("(%d, %d)\n",path[i].x,path[i].y);
		}
		return;
	}

	for(int i = 0; i < turn; i++)
	{
		int p = x + moves[i][0];
		int q = y + moves[i][1];
		if(p >= 1 && p <= n && q >= 1 && q <= m && !visited[p][q] && map[p][q] == 0)
		{
			visited[p][q] = true;
			back[p][q].x = x;
			back[p][q].y = y;
			dfs(p, q);
			if(flag) return;
			visited[p][q] = false;
		}
	}
}

void ShowSetp()
{
	HANDLE handle;
	handle = GetStdHandle(STD_OUTPUT_HANDLE);
	memset(color, 0, sizeof(color));
	for(int i = 0; i <= ans; i++)
	{
		printf("(%d, %d)\n",path[i].x,path[i].y);
		color[path[i].x][path[i].y] = 1;
	}
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= m; j++)
		{
			if(color[i][j] == 1)
			{
				SetConsoleTextAttribute(handle,  
										FOREGROUND_GREEN |  
										FOREGROUND_INTENSITY);  
				cout << map[i][j] << " ";
			}
			else
			{
				SetConsoleTextAttribute(handle,  
					FOREGROUND_RED |  
					FOREGROUND_INTENSITY);  
				cout << map[i][j] << " ";
			}
		}
		cout << endl;
	}
}
																	
int main()
{
	LoadMap();
	ShowMap();
	memset(visited, false, sizeof(visited));
	ans = 0;
	visited[1][1] = true;
	flag = false;
	dfs(1, 1);
	if(flag)
	{
		cout << "It will take " << ans << " setps to the destination" << endl;
		ShowSetp();
	}
	else
	{
		cout << "Can't Find a way to the destination" << endl;
	}
	return 0;
}

《模拟求解迷宫问题（DFS+BFS）》

话说，记录这张图片等了好久的说。结点一多，DFS的效率就降低了。

    原文作者：N3verL4nd
    原文地址: https://www.cnblogs.com/lgh1992314/archive/2013/04/17/5835113.html
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