迷宫问题 Maze 4X4 (sloved by backtrack)

 1 #include <iostream>
 2 #include <vector>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 typedef struct
 7 {
 8     int x;
 9     int y;
10 }pos;
11 
12 #define N 4  //4*4 maze
13 #define ENTER_X 0
14 #define ENTER_Y 0
15 #define EXIT_X N-1
16 #define EXIT_Y N-1
17 
18 int Maze[N][N];
19 int paths;  //sum of paths
20 vector<pos> Path;  //svae paths
21 vector<pos> BestPath; //save min distant path
22 
23 void InitMaze();
24 bool MazeTrack(int x,int y); 
25 
26 int main()
27 {
28     InitMaze();
29     int i = 0 , j = 0;
30     for(i=0;i<N;i++)
31     {    for(j=0;j<N;j++)
32             cout << Maze[i][j];
33         cout << endl;
34     }
35     MazeTrack(ENTER_X,ENTER_Y);
36     return 0;
37 }
38 
39 void InitMaze()
40 {
41     int    a[N][N] = {
42         {0,0,0,0},
43         {1,0,1,0},
44         {1,0,1,0},
45         {1,0,1,0}
46         };
47     memcpy(Maze,a,sizeof(a));
48     paths = 0;
49 }
50 
51 bool MazeTrack(int x,int y)
52 {
53     bool IsPath = false;
54     pos p;
55     p.x = x;
56     p.y = y;
57     //set range of maze‘s x/y
58     if(x<N && x>=0 && y<N && y>=0 && Maze[x][y]==0)
59     {
60         //make
61         Path.push_back(p);
62         Maze[x][y] = 1;  //if value is 1,you can‘t go there
63     
64         cout << "now,position is("<<x<<","<<y<<")" << endl;
65         //terminal
66         if(x==EXIT_X && y==EXIT_Y)
67         {
68             cout << "find a path" << endl;
69             paths++;
70             IsPath = true;
71             vector<pos>::iterator it;
72             for(it=Path.begin() ; it!=Path.end() ; it++)
73                 cout << "("<< it->x <<","<< it->y <<")" << " ";
74             cout << endl;
75             return IsPath;
76         }
77         //search  (find forward solutions)
78         IsPath = MazeTrack(x-1,y) || MazeTrack(x,y-1) || MazeTrack(x+1,y) || MazeTrack(x,y+1);
79         //backtrack
80         if(!IsPath)
81         {
82             Path.pop_back();  
83             Maze[x][y] = 0;
84         }
85         cout << Path.size() << endl;
86     }
87     //fuction exit
88     return IsPath;    
89 } 

 1 using namespace std;
 2 #define N 100  
 3 #define M 100
 4 
 5 typedef struct  
 6 {  
 7        int x;  
 8        int y;
 9 }Point; 
10  
11 vector<Point>solutionPath ;  //存放有解的坐标路径
12 
13 //主函数 x,y默认为起始结点，如(0,0), 得到从起始结点到目的结点的路径。
14 bool hasSolutionFunction( template<typename T>* Matrix , int x, int y)
15 {
16         
17     bool *visited = new bool[M*N]; 
18     memset (visited,0,M*N); //visited 存放每个结点是否被遍历，true为已经遍历过，false为否
19     
20     res =  hasSolutionCore(Matrix , x ,y)
21     cout<<solutionPath<<endl;
22     return res
23 
24 }
25 
26 //深度遍历求解路径的函数
27 bool hasSolutionCore(template<typename T>*  Matrix , int x, int y)
28 {
29     
30     hasSolution = false;
31     Point p = Point(x,y);
32     
33 
34         
35     if (x,y) is terminal  //x,y 已经是终止条件，当求解到这个结点是叶结点或目的地    
36         {
37         solutionPath.push_back(p);
38         return true;
39         }
40 
41     if ((x,y) && visited[x][y]==flase )// x,y是合法结点(具体条件可扩展)，且未被访问过
42     {
43         visited[x][y] = true;
44         solutionPath.push_back(p);
45         
46         hasSolution = hasSolutionCore(Matrix,x-1, y) ||hasSolutionCore(Matrix,x,y-1)||... //如果不是叶结点，则该路径是否有解取决于其他方向的往后求解。
47         
48         // x,y结点以下的所有子路径都无解，则回溯
49         if (!hasSolution)
50         {
51         visited[x][y] = false;
52         solutionPath.pop_back();
53         }
54         
55     }
56     return hasSolution;
57     
58 }