Children of the Candy Corn

Description

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

一道题，从学图到做出来，花了一天。

大意：

就是给出一个迷宫，然后按3种方式走出去：

     1.见岔路向左走。

     2.见岔路向右走。

     3.最短路。

第三种我最陌生难以理解，学了bfs,轻松搞定，1、2花了很大力气，实在不好写，因为要记录方向，然后判断，我利用了两个数组，画图演示一下

下面是代码，有点长，

出了一点解释的错误，下面代码中的所有注释的栈都是·队列

  1 #include<cstdio>

  2 #include<cstdlib>

  3 #include<cstring>

  4 #include<iostream>

  5 #include<queue>

  6 using namespace std;

  7 int s[50][50],vis[50][50];

  8 int si,sj,ei,ej;

  9 int r[4]= {0,0,1,-1};//这个是访问上下左右

 10 int c[4]= {1,-1,0,0};

 11 struct node

 12 {

 13     int i,j,step;

 14 } p,ph;

 15 int ri[12]= {1,0,-1,0,1,0,-1,0,1,0,-1,0};//这个是为了方便转方向，具体用途刚才解释过

 16 int rj[12]= {0,-1,0,1,0,-1,0,1,0,-1,0,1};

 17 queue<node>que;

 18 void get(int a,int b)

 19 {

 20     int i,j;

 21     char ch;

 22     for(i=1; i<a; i++)

 23     {

 24         for(j=1; j<b; j++)

 25         {

 26             scanf("%c",&ch);//这是一个简单地输入转换

 27             if(ch=='#')

 28                 s[i][j]=0;

 29             else

 30             {

 31                 if(ch=='S')

 32                 {

 33                     si=i;

 34                     sj=j;

 35                 }

 36                 if(ch=='E')

 37                 {

 38                     ei=i;

 39                     ej=j;

 40                 }

 41                 s[i][j]=1;

 42             }

 43         }

 44         getchar();//去掉换行符

 45     }

 46 }

 47 void bfs()//广度优先搜索，找最短路径

 48 {

 49     int k,a,b;

 50     p.i=si,p.j=sj,p.step=1;

 51     que.push(p);//入栈

 52     memset(vis,0,sizeof(vis));

 53     vis[si][sj]=1;

 54     while(!que.empty())

 55     {

 56         p=que.front();//取栈顶元素

 57         que.pop();//取后弹出

 58         for(k=0; k<4; k++)

 59         {

 60             a=p.i+r[k],b=p.j+c[k];

 61             if(a==ei&&b==ej)//是要找的就返回

 62             {

 63                 printf("%d\n",p.step+1);

 64                 return;

 65             }

 66             if(!vis[a][b]&&s[a][b])//如果未被访问过，又是可以可以走的路，就入栈

 67             {

 68                 ph.i=a,ph.j=b,ph.step=p.step+1;

 69                 que.push(ph);

 70                 vis[a][b]=1;

 71             }

 72         }

 73     }

 74 }

 75 void lr(char ch)//解决靠左靠右走的问题，合并了

 76 {

 77     int k,i,j,ia=si,ja=sj,ib,jb,step=1;

 78     for(k=0; k<4; k++)//第一步只有一种走法

 79     {

 80         if(s[ia+r[k]][ja+c[k]])

 81         {

 82             ib=ia+r[k];

 83             jb=ja+c[k];

 84             break;

 85         }

 86     }

 87     while(1)//直到找到正确的路再返回

 88     {

 89         if(ib==ei&&jb==ej)//找到正确的路返回

 90         {

 91             printf("%d ",++step);

 92             return ;

 93         }

 94         step++;

 95         i=ib-ia,j=jb-ja;

 96         ia=ib,ja=jb;

 97         for(k=3; k<8; k++)//没有的话，这里确定方向

 98             if(ri[k]==i&&rj[k]==j)

 99                 break;

100         if(ch=='R')//看是左优先还是右优先

101         {

102             k++;

103             if(s[ib+ri[k]][jb+rj[k]])//判断右侧是不是路，是就走

104             {

105                 ib+=ri[k],jb+=rj[k];

106                 continue;

107             }

108             for(k=k-1; k>0; k--)//不是依次向左找路

109                 if(s[ib+ri[k]][jb+rj[k]])

110                 {

111                     ib+=ri[k],jb+=rj[k];

112                     break;

113                 }

114         }

115         else//参考向右的

116         {

117             k--;

118             if(s[ib+ri[k]][jb+rj[k]])

119             {

120                 ib+=ri[k],jb+=rj[k];

121                 continue;

122             }

123             for(k=k+1;; k++)

124                 if(s[ib+ri[k]][jb+rj[k]])

125                 {

126                     ib+=ri[k],jb+=rj[k];

127                     break;

128                 }

129         }

130     }

131 }

132 int main()

133 {

134     int p,a,b;

135     cin>>p;

136     getchar();

137     while(p--)

138     {

139         memset(s,0,sizeof(s));

140         cin>>b>>a;

141         getchar();//输入不解释，其实主函数最简单 不解释了

142         get(a+1,b+1);

143         lr('L');

144         lr('R');

145         bfs();

146     }

147     return 0;

148 }

View Code