POJ 3083 Children of the Candy Corn (BFS ,DFS) -- from lanshui_Yang
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
题目大意：给你一张图，“#”代表 墙， “.”表示路， “S”和“E”分别代表起点，出口。有三种走法：左转优先，右转优先，正常走法。让你求出分别三种走法从起点走到出口所需的最小步数（其中，起点“S” 算是第一步）。显然，求正常走法的最短路径时，直接用BFS就行啦， 但是此题难点在于如何应用DFS求出 左转优先和右转优先 的最短路径 ，而DFS中的难点在于每次搜索时的前进方向应如何确定？？ 下面以 左转优先为例，向大家讲述。
先看一张图 ：
北（1）
西（0） 当前位置 东（2）
南（3）
如果路人甲 从 东（编号为2）来 到“当前位置 ”，那么此时 他 的前方是 西（编号为0） 并记录为 d = 0，他的 左手边是 南（编号 为 3），因为左转优先，于是 在走下一步时，他的 方向选择顺序依次是 3,0,1,2， 如果3能走，那么他下次在“当前位置”时 面朝的方向依旧是 3，同理，如果3走不通 ，2 能走通，那么路人甲 下次 在“当前位置”时 面朝的方向依旧是 2， 这样就方便记录啦！运用DFS 的关键 就在于此 ！！！
具体讲解请看程序：
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<string> #include<queue> #include<cmath> using namespace std; struct point { int x; int y; int dis; }; queue<point>q; int ldis ; // 定义左转优先的步数 int rdis ; // 定义右转优先的步数 char map[46][46]; int vis[46][46]; // 标记数组 ，用于BFS int X[4]= {0,-1,0,1}; // 定义前进方向，注意此处 方向 不是 随便定义的, //与上述解释中的各个方向的编号有关, int Y[4]= {-1,0,1,0}; point start,end; int n ,m; int di ,dj; // 终点坐标 int bfs(); int cango(int i,int j); // bfs 中能够前进的判断条件 int cangoDfs(int i,int j); // 两个dfs中能够前进的判断条件 void leftdfs(int si,int sj,int d); // 左转优先的dfs，其中，si，sj是起点坐标，d是当前位置面朝的方向~~ void rightdfs(int si,int sj,int d); // 右转优先的dfs int pan ; // 判断变量 ，dfs 中要用 ~ int main() { int t; scanf("%d",&t); while (t--) { scanf("%d%d",&m,&n); int si,sj,d; int i,j; for(i=1; i<=n; i++) { for(j=1; j<=m; j++) { cin>>map[i][j]; if(map[i][j]=='S') { si = i; sj = j; start.x = i; start.y = j; if(map[si][sj-1]!='#') // 判断路人甲在起点”S“时，面朝的方向 ！！ d = 0; if(map[si-1][sj]!='#') d = 1; if(map[si][sj+1]!='#') d = 2; if(map[si+1][sj]!='#') d = 3; } if(map[i][j]=='E') { di = i; dj = j; end.x = i; end.y = j; } } } ldis = rdis = 1; // 把左转优先和右转优先的步数 均初始化为 1（因为'S'（即起点）本身算是第一步！）~ pan = 0; // 把判断变量 赋为 0 leftdfs(si,sj,d); pan = 0; rightdfs(si,sj,d); while (!q.empty()) // 调用 队列 前 先清空（此处必须有，原因在下面的bfs中解释） q.pop(); memset(vis,0,sizeof(vis)); // 标记数组全部初始化为 0 int sd = bfs(); printf("%d %d %d\n",ldis,rdis,sd); } return 0; } int bfs() { start.dis = 1; vis[start.x][start.y] = 1; q.push(start); point hd,next; while (!q.empty()) { hd = q.front(); q.pop(); int k; for(k = 0; k < 4; k++) { next.x = hd.x + X[k]; next.y = hd.y + Y[k]; if(cango(next.x,next.y)) { next.dis = hd.dis + 1; if(next.x == end.x && next.y == end.y) { return next.dis; // 因为此处直接return ，剩下的队列中可能还有元素， //所以每次调用队列前都要先清空 } vis[next.x][next.y] = 1; // 访问过的点标记为 1 q.push(next); } } } } // 此题的dfs中 不能标记数组，因为可能要回退！！ void leftdfs(int si,int sj,int d) { int k; for(k=0; k<4; k++) { if(pan ==1) // 此处 也很关键 ，如果没有 会 死循环 ！！！ return ; int tmp = (d + 3 + k)%4; // 此处 是 每次前进时方向与 k 的关系(请大家动手，很好推~) if(cangoDfs(si+X[tmp],sj+Y[tmp])) { ldis ++; if(si+X[tmp]==di && sj+Y[tmp] == dj) { pan = 1; // 如果找到 终点 就把 pan 的值 变为 1，然后return return ; } leftdfs(si+X[tmp],sj+Y[tmp],tmp); } } return ; } void rightdfs(int si,int sj,int d) { int k; for(k=0; k<4; k++) { if(pan ==1) return ; int tmp = (d + 5 - k)%4; // 此处 是 每次前进时方向与 k 的关系 if(cangoDfs(si+X[tmp],sj+Y[tmp])) { rdis ++; if(si+X[tmp]==di && sj+Y[tmp] == dj) { pan = 1; return ; } rightdfs(si+X[tmp],sj+Y[tmp],tmp); } } return ; } int cango(int i,int j) { if(map[i][j]!='#'&&i>=1&&i<=n&&j>=1&&j<=m&&vis[i][j]==0) return 1; return 0; } int cangoDfs(int i,int j) { if(map[i][j]!='#'&&i>=1&&i<=n&&j>=1&&j<=m) return 1; return 0; }