The cornfield maze is a popular Halloween treat. Visitors are shown the entrance and must wander through the maze facing zombies, chainsaw-wielding psychopaths, hippies, and other terrors on their quest to find the exit. 

One popular maze-walking strategy guarantees that the visitor will eventually find the exit. Simply choose either the right or left wall, and follow it. Of course, there's no guarantee which strategy (left or right) will be better, and the path taken is seldom the most efficient. (It also doesn't work on mazes with exits that are not on the edge; those types of mazes are not represented in this problem.) 

As the proprieter of a cornfield that is about to be converted into a maze, you'd like to have a computer program that can determine the left and right-hand paths along with the shortest path so that you can figure out which layout has the best chance of confounding visitors.

Input

Input to this problem will begin with a line containing a single integer n indicating the number of mazes. Each maze will consist of one line with a width, w, and height, h (3 <= w, h <= 40), followed by h lines of w characters each that represent the maze layout. Walls are represented by hash marks ('#'), empty space by periods ('.'), the start by an 'S' and the exit by an 'E'. 

Exactly one 'S' and one 'E' will be present in the maze, and they will always be located along one of the maze edges and never in a corner. The maze will be fully enclosed by walls ('#'), with the only openings being the 'S' and 'E'. The 'S' and 'E' will also be separated by at least one wall ('#'). 

You may assume that the maze exit is always reachable from the start point.

Output

For each maze in the input, output on a single line the number of (not necessarily unique) squares that a person would visit (including the 'S' and 'E') for (in order) the left, right, and shortest paths, separated by a single space each. Movement from one square to another is only allowed in the horizontal or vertical direction; movement along the diagonals is not allowed.

Sample Input

2
8 8
########
#......#
#.####.#
#.####.#
#.####.#
#.####.#
#...#..#
#S#E####
9 5
#########
#.#.#.#.#
S.......E
#.#.#.#.#
#########

Sample Output

37 5 5
17 17 9

POJ3083

題意：

先沿著左邊的牆從 S 一直走，求到達 E 的步數。

再沿著右邊的牆從 S 一直走，求到達 E 的步數。

最後求最短路。

解題思路：

最短路好辦，關鍵是沿著牆走不太好想。

但只要弄懂如何轉，這題就容易了。

單就沿著左走看一下：

當前方向 　　  檢索順序

     ↑ ：    　　← ↑ → ↓

    → ：    　   ↑ → ↓ ← 

     ↓ ：    　　→ ↓ ← ↑ 

    ← ：    　   ↓ ← ↑ → 

如此，規律很明顯，假設陣列存放方向為 ← ↑ → ↓， 如果當前方向為 ↑， 就從 ← 開始依次遍歷，找到可以走的，如果 ← 可以走，就不用再看 ↑ 了。

在DFS時，加一個引數，用來儲存當前的方向。

AC程式碼：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

char mp[110][110];
int sx,sy,ex,ey,n,m,book[110][110];

int nx[4][2]={0,-1,-1,0,0,1,1,0};//bfs
int nxl[4][2]={0,-1,-1,0,0,1,1,0};//ldfs
int nxr[4][2]={0,1,-1,0,0,-1,1,0};//rdfs

struct node
{
	int x,y,step;
};

node getnode(int x,int y,int step)
{
	node q;
	q.x=x;
	q.y=y;
	q.step=step;
	return q;
}

int dfs(int x,int y,int d,int step,int dir[][2])
{
	for(int i=0;i<4;i++)
	{
		int j=((d-1+4)%4+i)%4;
		int tx=x+dir[j][0];
		int ty=y+dir[j][1];
		if(tx>=0&&tx<n&&ty>=0&&ty<m&&mp[tx][ty]!='#')
		{
			if(tx==ex&&ty==ey) return step+1;
			return dfs(tx,ty,j,step+1,dir);
		}
	}
}

int bfs(int sx,int sy)
{
	queue<node> q;
	q.push(getnode(sx,sy,1));
	while(!q.empty())
	{
		for(int i=0;i<4;i++)
		{
			int tx=q.front().x+nx[i][0];
			int ty=q.front().y+nx[i][1];
			if(tx>=0&&tx<n&&ty>=0&&ty<m&&book[tx][ty]==0&&mp[tx][ty]!='#')
			{
				book[tx][ty]=1;
				if(tx==ex&&ty==ey) return q.front().step+1;
				q.push(getnode(tx,ty,q.front().step+1));
			}
		}
		q.pop();
	}
	return -1;
}

int main()
{
	int t,d1,d2;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&m,&n);
		for(int i=0;i<n;i++)
			scanf("%s",mp[i]);
			
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				if(mp[i][j]=='S') {sx=i;sy=j;}
				else if(mp[i][j]=='E') 	{ex=i;ey=j;}
			}
		}
		
		if(sx==0) {d1=3;d2=3;}
		else if(sx==n-1) {d1=1;d2=1;}
		else if(sy==0) {d1=2;d2=0;}
		else if(sy==m-1) {d1=0;d2=2;}
		
		memset(book,0,sizeof(book));
		book[sx][sy]=1;
		
		printf("%d ",dfs(sx,sy,d1,1,nxl));
		printf("%d ",dfs(sx,sy,d2,1,nxr));
		printf("%d\n",bfs(sx,sy));
	}
	return 0;//
}