Children of the Candy Corn
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6630 | Accepted: 2891 |
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
转自：http://www.cnblogs.com/lyy289065406/archive/2011/07/31/2122369.html
题目大意:
给定一个迷宫，S是起点，E是终点，#是墙不可走，.可以走
先输出左转优先时，从S到E的步数
再输出右转优先时，从S到E的步数
最后输出S到E的最短步数
W为宽，列数
H为高，行数
解题思路：
DFS和BFS的综合题水题，难度不大，但是写代码时要注意几方面：
1、 左转、右转优先搜索时必须标记当前位置时的方向，我定义的方向是
最初的方向由起点S确定，而下一步的方向则由前一步的走向决定
例如 左边优先搜索：
当前位置的方向指向 1（向左），(这同时说明前一步是在第“3”的位置走过来的)
那么走下一步时，就要根据2103的顺序，先逐格确定当前位置周边的四格是否可行
若第一次确认2可行，就走到2，在位置2时的方向为2（向下）
若2不可行，则再确定1，若1可行，就走到1，在位置1时的方向为1（向左）
若1也不可行，则再确定0，若0可行，就走到0，在位置0时的方向为0（向上）
若0也不可行，说明进入了迷宫的死胡同，要从原路返回，走回3
右边优先搜索也同理。
根据我定义的方向，设当前位置为d，那么
左转，用数学式子表达就是 d=(d+1)%4
右转，用数学式子表达就是 d=(d+3)%4
我比较懒，在我的程序中，DFS和BFS都用了多入口的做法，有兴趣的同学可以利用我给出的这两个式子对代码进行优化。
这里有一点必须要注意的：
左边、右边优先搜索都不是找最短路，因此走过的路可以再走，无需标记走过的格
2、 寻找最短路只能用BFS
因此在做第3问时别傻乎乎的又用DFS，DFS对于样例的输入确实和BFS得到的结果一样的，别以为样例PASS就提交了。。。所以我就说样例没代表性，学会测试数据很重要= =
注意有一点：
要求E的最短路，必须把迷宫模拟为树，S为根，找到E所在的层（树深），该层就是S到E的最短路，处理技巧就是在BFS时，令queue[tail]的depth等于对应的queue[head]的depth+1，详细见我的程序
把循环的次数作为深度就铁定错的
代码：
View Code
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 6 struct M{ 7 int r,c; 8 int depth; 9 }; 10 11 M s,e; 12 int Lstep,Rstep; //左/右优先的步数 13 14 bool maze[41][41]; 15 16 void DFS_L(int i,int j,int d) //左边优先搜索， i,j是当前坐标，d为位置方向 17 { 18 Lstep++; 19 if(i==e.r && j==e.c) 20 { 21 return ; 22 } 23 switch(d) //由图可以看出 24 { 25 case 0: 26 { 27 if(maze[i][j-1]) 28 DFS_L(i,j-1,1); 29 else if(maze[i-1][j]) 30 DFS_L(i-1,j,0); 31 else if(maze[i][j+1]) 32 DFS_L(i,j+1,3); 33 else if(maze[i+1][j]) 34 DFS_L(i+1,j,2); 35 break; 36 } 37 case 1: 38 { 39 if(maze[i+1][j]) 40 DFS_L(i+1,j,2); 41 else if(maze[i][j-1]) 42 DFS_L(i,j-1,1); 43 else if(maze[i-1][j]) 44 DFS_L(i-1,j,0); 45 else if(maze[i][j+1]) 46 DFS_L(i,j+1,3); 47 break; 48 } 49 case 2: 50 { 51 if(maze[i][j+1]) 52 DFS_L(i,j+1,3); 53 else if(maze[i+1][j]) 54 DFS_L(i+1,j,2); 55 else if(maze[i][j-1]) 56 DFS_L(i,j-1,1); 57 else if(maze[i-1][j]) 58 DFS_L(i-1,j,0); 59 break; 60 } 61 case 3: 62 { 63 if(maze[i-1][j]) 64 DFS_L(i-1,j,0); 65 else if(maze[i][j+1]) 66 DFS_L(i,j+1,3); 67 else if(maze[i+1][j]) 68 DFS_L(i+1,j,2); 69 else if(maze[i][j-1]) 70 DFS_L(i,j-1,1); 71 break; 72 } 73 } 74 return ; 75 } 76 77 void DFS_R(int i,int j,int d) 78 { 79 Rstep++; 80 if(i==e.r && j==e.c) 81 return; 82 switch(d) 83 { 84 case 0: 85 { 86 if(maze[i][j+1]) 87 DFS_R(i,j+1,3); 88 else if(maze[i-1][j]) 89 DFS_R(i-1,j,0); 90 else if(maze[i][j-1]) 91 DFS_R(i,j-1,1); 92 else if(maze[i+1][j]) 93 DFS_R(i+1,j,2); 94 break; 95 } 96 case 1: 97 { 98 if(maze[i-1][j]) 99 DFS_R(i-1,j,0); 100 else if(maze[i][j-1]) 101 DFS_R(i,j-1,1); 102 else if(maze[i+1][j]) 103 DFS_R(i+1,j,2); 104 else if(maze[i][j+1]) 105 DFS_R(i,j+1,3); 106 break; 107 } 108 case 2: 109 { 110 if(maze[i][j-1]) 111 DFS_R(i,j-1,1); 112 else if(maze[i+1][j]) 113 DFS_R(i+1,j,2); 114 else if(maze[i][j+1]) 115 DFS_R(i,j+1,3); 116 else if(maze[i-1][j]) 117 DFS_R(i-1,j,0); 118 break; 119 } 120 case 3: 121 { 122 if(maze[i+1][j]) 123 DFS_R(i+1,j,2); 124 else if(maze[i][j+1]) 125 DFS_R(i,j+1,3); 126 else if(maze[i-1][j]) 127 DFS_R(i-1,j,0); 128 else if(maze[i][j-1]) 129 DFS_R(i,j-1,1); 130 break; 131 } 132 } 133 return; 134 } 135 136 void BFS(int i, int j) // i,j为起始位置 137 { 138 bool vis[41][41]={false}; 139 M queue[1605]; 140 int head,tail; 141 head=tail=0; 142 queue[0].r=i; 143 queue[0].c=j; 144 queue[tail++].depth=1; 145 vis[i][j]=true; 146 147 while(head<tail) 148 { 149 M x=queue[head++]; 150 if(x.r==e.r && x.c==e.c) 151 { 152 printf("%d\n",x.depth); 153 return; 154 } 155 if(maze[x.r][x.c-1] && !vis[x.r][x.c-1]) 156 { 157 vis[x.r][x.c-1]=true; 158 queue[tail].r=x.r; 159 queue[tail].c=x.c-1; 160 queue[tail++].depth=x.depth+1; 161 } 162 if(maze[x.r-1][x.c] && !vis[x.r-1][x.c]) 163 { 164 vis[x.r-1][x.c]=true; 165 queue[tail].r=x.r-1; 166 queue[tail].c=x.c; 167 queue[tail++].depth=x.depth+1; 168 } 169 if(maze[x.r][x.c+1] && !vis[x.r][x.c+1]) 170 { 171 vis[x.r][x.c+1]=true; 172 queue[tail].r=x.r; 173 queue[tail].c=x.c+1; 174 queue[tail++].depth=x.depth+1; 175 } 176 if(maze[x.r+1][x.c] && !vis[x.r+1][x.c]) 177 { 178 vis[x.r+1][x.c]=true; 179 queue[tail].r=x.r+1; 180 queue[tail].c=x.c; 181 queue[tail++].depth=x.depth+1; 182 } 183 184 } 185 return; 186 } 187 188 int main() 189 { 190 int t; 191 int i,j; 192 scanf("%d",&t); 193 while(t--) 194 { 195 int dir; 196 int w,h; 197 scanf("%d%d",&w,&h); 198 Lstep=1; 199 Rstep=1; 200 memset(maze,false,sizeof(maze)); 201 202 for(i=1;i<=h;i++) 203 { 204 for(j=1;j<=w;j++) 205 { 206 char tt; 207 cin>>tt; 208 if(tt=='.') 209 maze[i][j]=true; 210 if(tt=='S') 211 { 212 maze[i][j]=true; 213 s.r=i; 214 s.c=j; 215 if(i==h) dir=0; //判断其实dir方向 216 else if(j==w) dir=1; 217 else if(i==1) dir=2; 218 else if(j=1) dir=3; 219 } 220 if(tt=='E') 221 { 222 maze[i][j]=true; 223 e.r=i; 224 e.c=j; 225 } 226 227 } 228 } 229 switch(dir) 230 { 231 case 0: {DFS_L(s.r-1,s.c,0);break;} // 题目描述，所有S两边都是'#'。 232 case 1: {DFS_L(s.r,s.c-1,1);break;} 233 case 2: {DFS_L(s.r+1,s.c,2);break;} 234 case 3: {DFS_L(s.r,s.c+1,3);break;} 235 } 236 cout<<Lstep<<' '; 237 238 switch(dir) 239 { 240 case 0: {DFS_R(s.r-1,s.c,0);break;} 241 case 1: {DFS_R(s.r,s.c-1,1);break;} 242 case 2: {DFS_R(s.r+1,s.c,2);break;} 243 case 3: {DFS_R(s.r,s.c+1,3);break;} 244 } 245 cout<<Rstep<<' '; 246 247 BFS(s.r,s.c); 248 } 249 return 0; 250 }
