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.
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.
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 ",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 }