P1605 maze, p1605 maze

P1605 maze, p1605 maze
Background

Maze [Problem description]

Given a maze of N * M squares, there are T obstacles in the Maze and obstacles cannot pass through. Given the start coordinate and

Endpoint coordinate: How many methods can each square go through at most one time from the start point coordinate to the end point coordinate. In the maze

China Mobile has four ways to move up, down, and left. Only one square can be moved at a time. There are no barriers to data assurance.

Input sample output sample

[Data scale]

1 ≤ N, M ≤ 5

Question description input/output format Input Format:

[Input]

The first row N, M, and T, N is the row, M is the column, and T is the total number of obstacles. Second line starting point coordinate SX, SY, end point

Coordinate FX, FY. The coordinates of each behavior obstacle point in the next T line.

Output Format:

[Output]

Given the start coordinate and end coordinate, ask each square that passes through a maximum of one time, from the start coordinate to the end Coordinate

Total number of cases.

Input and Output sample Input example #1:

2 2 11 1 2 21 2

Output sample #1:

1

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 int bgx,bgy,edx,edy; 7 int n,m,t; 8 int xx[5]={-1,+1,0,0}; 9 int yy[5]={0,0,-1,+1};10 int map[50][50];11 int ans=0;12 int vis[50][50];13 void dfs(int x,int y)14 {15     if(x==edx&&y==edy)16     {17         ans++;18         return ;19     }20     for(int i=0;i<4;i++)21     {22         int wx=x+xx[i];23         int wy=y+yy[i];24         if(vis[wx][wy]==0&&map[wx][wy]!=1&&wx>=1&&wx<=n&&wy>=1&&wy<=m)25         {26             vis[wx][wy]=1;27             dfs(wx,wy);28             vis[wx][wy]=0;29         }30     }31     32 }33 int main()34 {35     36     scanf("%d%d%d%d%d%d%d",&n,&m,&t,&bgx,&bgy,&edx,&edy);37     for(int i=1;i<=t;i++)38     {39         int x,y;40         scanf("%d%d",&x,&y);41         map[x][y]=1;42     }43     vis[bgx][bgy]=1;44     dfs(bgx,bgy);45     printf("%d",ans);46     return 0;47 }