WalkingMachine

Given a maze of size $n\times m$ , where upper left corner is $(1, 1)_{}$ and lower right corner is $(n, m)_{}$ . For each cell $(x, y)_{}$ , there is exactly one character $c~(c\in \{W, A, S, D\})$ on it, denoting the moving restriction:

* If $c = W_{}$ , you can only go up from this cell, specifically go from $(x, y)_{}$ to $(x-1, y)_{}$

* If $c = A_{}$ , you can only go left from this cell, specifically go from $(x, y)_{}$ to $(x, y-1)_{}$

* If $c = S_{}$ , you can only go down from this cell, specifically go from $(x, y)_{}$ to $(x+1, y)_{}$

* If $c = D_{}$ , you can only go right from this cell, specifically go from $(x, y)_{}$ to $(x, y+1)_{}$

We say one is out if $x < 1_{}$ or $x > n_{}$ or $y < 1_{}$ or $y > m_{}$ holds, now you should determine the number of cells that one can be out if start walking on it.

#include<cstdio> int n,m,ans,vis[1005][1005];char s[1005][1005]; int work(int x,int y) { if(x<1||x>n||y<1||y>m)return 1; if(vis[x][y])return vis[x][y]; vis[x][y]=2;int xx=x,yy=y; if(s[x][y]=='W')x--;else if(s[x][y]=='S')x++;else if(s[x][y]=='A')y--;else if(s[x][y]=='D')y++; return vis[xx][yy]=work(x,y); } int main() { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++)scanf("%s",s[i]+1); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { if(!vis[i][j])work(i,j); if(vis[i][j]==1)ans++; } printf("%d\n",ans); }

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