POJ 3083：Children of the Candy Corn（DFS+BFS）

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

1.题意：有一个迷宫，#代表墙，..代表能走，S是起点，E是终点W为宽，列数H为高；

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

自己写的有很多问题。。后面我发现别人都是用什么数学方法来确定向左还是向右。。我马上就Orz了。。

那些人里面，写的最好的就是这个了点击打开链接。。尼玛，又看了结题报告。。╮(╯▽╰)╭。。。

简直丧心病狂。。剁手。。。好吧。。题外话就不多说了。。。其他的他都说的很详细了。。。我也就

打打酱油吧。。。。。。。。Orz。。。。。。。。。。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>

using namespace std;

const int N = 105;

char map[N][N];
int vist[N][N];

struct node
{
    int x;
    int y;
    int num;
};
queue<node>q;
node  first;

int dx[4]={1,-1,0,0};
int dy[4]={0,0,-1,1};
int fx[]= {0,1,0,-1};
int fy[]= {1,0,-1,0};
int fr[]= {1,0,3,2};
int fl[]= {3,0,1,2};
int ans;
int t, n, m;
int xx, yy;
int d;

void L_dfs(int x, int y, int d)  //靠左墙
{
    ans++;
    if( map[x][y] == 'E' )
    {
        printf( "%d", ans );
        ans = 0;     //记得初始
        return ;
    }
    for(int i=0; i<4; i++)
    {
        int j = ( d + fl[i] ) % 4;
        xx = x + fx[j];
        yy = y + fy[j];
        if(xx>=1 && xx<=n && yy>=1 && yy<=m && map[xx][yy]!='#')
        {
            L_dfs(xx, yy, j);
            return ;   //少了直接爆掉
        }
    }

}

void R_dfs(int x, int y, int d)  //向右
{
    ans++;
    if( map[x][y] == 'E' )
    {
        printf(" %d", ans );
        ans = 0;
        return ;
    }
    for(int i=0; i<4; i++)
    {
        int j= ( d + fr[i] ) % 4;
        xx = x + fx[j];
        yy = y + fy[j];
        if(xx>=1 && xx<=n && yy>=1 && yy<=m && map[xx][yy]!='#')
        {
            R_dfs(xx, yy, j);
            return ;
        }
    }

}

void S_bfs()  //最短路径
{
    memset( vist, false, sizeof( vist ) );
    vist[first.x][first.y] = true;
    while( !q.empty() )
    {
        node temp = q.front();
        q.pop();
        if( map[temp.x][temp.y]=='E' )
        {
            printf(" %d\n", temp.num+1);
            break;
        }
        for(int i=0; i<4; i++)
        {

            xx = temp.x + dx[i];
            yy = temp.y + dy[i];
            if( xx>=1 && xx<=n &&yy>=1 &&yy<=m && !vist[xx][yy] && map[xx][yy]!='#' )
            {
                node next;
                next.x = xx;
                next.y = yy;
                next.num = temp.num + 1;
                vist[xx][yy] = true;
                q.push( next );
            }
        }
    }
}

int main()
{
    scanf("%d\n", &t);
    while( t-- )
    {
        memset( vist, false, sizeof( vist ) );
        while( !q.empty() )   q.pop();
        scanf("%d%d", &m, &n);
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
            {
                cin>>map[i][j];
                if( map[i][j]=='S' )
                {
                    first.x = i;
                    first.y = j;
                }
            }
        first.num = 0; ans = 0;
        vist[first.x][first.y] = true;
        q.push( first );
        if(first.x==1)     d=0;
        if(first.x==n)     d=2;
        if(first.y==1)     d=1;
        if(first.y==m)     d=3;
        L_dfs( first.x, first.y, d);
        R_dfs( first.x, first.y, d);
        S_bfs();
    }

    return 0;
}

POJ 3083：Children of the Candy Corn（DFS+BFS）,布布扣,bubuko.com