POJ 3083 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

虽然都说水题，但是真的感觉左遍历和右遍历的时候处理起来很复杂，代码也显得十分冗长。

#include 
#include 
int next[4][2]={{0,-1},{-1,0},{0,1},{1,0}},book[50][50];//
int u=1,d=3,r=2,l=0,min=9999999;// 主要是方向的建立比较重要 
char map[50][50];
struct node{//一开始打算将四个方向都标记出来，其实只需要正前方一个方向转向就好了 
	int f;
	int h;
	int r;
	int l;
}dir;
struct nodx{
	int x;
	int y;
}s,e;
struct kk{
	int x,y,step;
}que[100100],qc,qu;
int head,end;
int m,n;
void bfs()
{
	int i,j;
	while (head=m||y>=n)	continue;
			if (!book[x][y]&&map[x][y]!='#'){
				book[x][y]=1;
				que[end].x=x;
				que[end].y=y;
				que[end].step=que[head].step+1;
				if (x==e.x&&y==e.y){
					min=que[end].step;
					return ;
				}
				end++;
			}
		}
		head++;
	}
}
int  main ()
{
	int t,i;
	scanf ("%d",&t);
	while(t--){
		scanf ("%d%d",&m,&n);
		i=m;m=n;n=i;
		getchar ();
		int step_r=1,step_l=1,j;
		for (i=0;i=0&&x+next[d][0]=0&&y+next[d][1]=m&&y+next[d][1]<0&&y+next[d][1]>=n)	continue;
					if (map[x+next[d][0]][y+next[d][1]]!='#'){
						x+=next[d][0];y+=next[d][1];
						step_l++;
					//	printf ("%d-%d\n",x,y);
						break;
					}
					if (tt==3)	break;
				}
			}
		}
		x=s.x;y=s.y;d=dir.f;
		while (x!=e.x||y!=e.y){
			int tt=0;
			d++;//右遍历同理右转一次 
			if (d==4)	d=0;
			if ((x+next[d][0]>=0&&x+next[d][0]=0&&y+next[d][1]=m&&y+next[d][1]<0&&y+next[d][1]>=n)	continue;
					if (map[x+next[d][0]][y+next[d][1]]!='#'){
						x+=next[d][0];y+=next[d][1];
						step_r++;
					//	printf ("%d-%d\n",x,y);
						break;
					}
					if (tt==3)	break;
				}
			}
		}
		printf ("%d %d %d\n",step_l,step_r,min);
		memset(book,0,sizeof(book));
		memset(map,0,sizeof(map));
		memset(que,0,sizeof(que));
		min=999999;step_l=1;step_r=1;
	}
	return 0;
}

以上代码很多地方是不必要的比如min=9999999是之前用深搜超时后没有改正过来。

下面贴一个别人的左右遍历的方法

int Depth_FirstSearch(int x1,int y1,int d,int x2,int y2)  
{  
    int new_x,new_y,i,count=1;  
    if(x1==x2&&y1==y2)  
        return count;  
    d=(d+3)%4;  
    i=d;  
    for(i=d;i=0&&new_x=0&&new_y

形参依次是起始坐标，方向，结束坐标。d=(d+3)%4;代表左转90°。所以这个函数只能实现左遍历。然而-.-

        count1=Depth_FirstSearch(start_x,start_y,0,end_x,end_y); //左遍历
        count2=Depth_FirstSearch(end_x,end_y,0,start_x,start_y);//右遍历

右遍历正好等于终点到起点的左遍历。。。利用这个规律便极大的优化了代码量。