POJ3083

http://sfiction.blog.163.com/blog/static/19940401020124355831319/  感谢作者的思路拯救了我

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

//挖掘机_Coder_zha学习之路
#include <stdio.h>
void Init();
int Go(int k);
int BFS();
int a[40][40],d[4][2]={{-1,0},{0,-1},{1,0},{0,1}};
int e[2][8][2]={{{0,-1},{-1,0},{0,1},{1,0},{0,-1},{-1,0},{0,1},{1,0}},{{0,1},{-1,0},{0,-1},{1,0},{0,1},{-1,0},{0,-1},{1,0}}};
//第一组是左转的坐标加法，第二组是右转的坐标加法。 
int n,m,x0,y0,x1,y1;
void Init()
{
    int i,j;
    char chr;
    for (i=0;i<n;i++)
        for (getchar(),j=0;j<m;j++)
        {
            scanf("%c",&chr);
            if (chr=='#') a[i][j]=-1;
            else
            {
                if (chr=='S') {x0=i;y0=j;}
                else if (chr=='E') {x1=i;y1=j;}
                a[i][j]=0;
            }
        }
}
int Go(int k)//把两个方向的搜索合并成了一个函数，以下注释以向左为例//用k的值来控制左转，右转。 
{
    int d,i,x,y,x2,y2,s=1;
    a[x1][y1]=2;
    d=0;//初始方向无所谓，因为出口周围四个方向只有一个是空地，第一次行走必然会调整好
    x=x0;
    y=y0;
    while (a[x][y]==0)
        for (d=(d+3)%4,i=d;i<d+4;i++)//以左转90°后的方向为第一优先，依次右转90°找可能的通路
        {
            x2=x+e[k][i][0];//目标格子
            y2=y+e[k][i][1];
            if (x2>=0&&y2>=0&&x2<n&&y2<m&&a[x2][y2]>=0)//边界即空地判定
            {
                x=x2;
                y=y2;
                d=i%4;//设定新方向？？？？？？ 
                s++;//记录步数
                break;//完成一次。 
            }
        }
    return s;
}
int BFS()//宽搜解决最短路径
{
    int c[1600][2];
    int i,j,t0,t1,t2,t3,x,y;
    a[x1][y1]=0;
    c[0][0]=x0;
    c[0][1]=y0;
    a[x0][y0]=1;
    t1=0;
    t2=t3=1;
    t0=2;
    for (t1=0,t2=t3=1;t1<t2;t1=t2,t2=t3,t0++)
        for (i=t1;i<t2;i++)
        {
            for (j=0;j<4;j++)
            {
                x=c[i][0]+d[j][0];
                y=c[i][1]+d[j][1];
                if (x>=0&&y>=0&&x<n&&y<m&&a[x][y]==0)
                {
                    c[t3][0]=x;
                    c[t3][1]=y;
                    a[x][y]=t0;
                    t3++;
                }
            }
            if (a[x1][y1]>0) return a[x1][y1];
        }
    return 0;
}
int main()
{
    int n1;
    scanf("%d",&n1);
    for (;n1>0;n1--)
    {
        scanf("%d%d",&m,&n);
        Init();
        printf("%d ",Go(0));
        printf("%d ",Go(1));
        printf("%d\n",BFS());
    }
    return 0;
}

bfs好理解，就是原作者的转向搜索有点喵，不错不错，学习到了