ACM: 方向dfs + bfs求最短路 poj 3…

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 Inpu

2 8 8

########

# .  .  . 
.  .  . #

# . #### .#

#.####.#

#.####.#

#.####.#

#...#..#

#S#E####

9 5

#########

#.#.#.#.#

S.......E

#.#.#.#.#

#########

Sample Output

37 5 5

17 17 9

    

题意: 寻找路径. 从左搜 , 从右搜 , 求最短路径.

    

解题思路:

        
1. 最短路径用广搜.

        
2. 从左搜和从右搜,方向比较难处理. 规定dir[4][2] = { {0,1} , {1,0} , {0,-1} , {-1,0}
}; (东南西北)

        
3. 起点规定一定是在迷宫的边缘, 只要确定迷宫的4条边的情况.

 

代码:

 

#include <cstdio>

#include <iostream>

#include <cstring>

#include <queue>

using namespace std;

#define MAX 50

int dir[4][2] = { {0,1} , {1,0} , {0,-1} , {-1,0} };

int n , m;

char map[MAX][MAX];

bool visit[MAX][MAX];

struct node

{

    int x ,
y;

    int
dist;

};

int bfs(node start)

{

  
 memset(visit,false,sizeof(visit));

  
 visit[start.x][start.y] = true;

    node
t;

  
 queue<node>
qu;

  
 qu.push(start);

    int
result;

    

    while(
!qu.empty() )

    {

  
   
 t = qu.front();

  
   
 qu.pop();

  
   
 if(map[t.x][t.y] == 'E')

  
   
 {

  
   
   
 result = t.dist;

  
   
 }

  
   
 

  
   
 for(int i = 0; i <4; ++i)

  
   
 {

  
   
   
 node tt;

  
   
   
 tt.x = t.x + dir[i][0];

  
   
   
 tt.y = t.y + dir[i][1];

  
   
   
 tt.dist = t.dist + 1;

  
   
   
 if(tt.x >= 0
&& tt.x < n
&& tt.y >= 0
&& tt.y < m
&& !visit[tt.x][tt.y]
&& map[tt.x][tt.y] != '#')

  
   
   
 {

  
   
   
   
 visit[tt.x][tt.y] = true;

  
   
   
   
 qu.push(tt);

  
   
   
 }

  
   
 }

    }

    

    return
result;

}

int dfs(int sx,int sy,int direction,int flag)

{

    int
newdir;

    

  
 if(map[sx][sy] == 'E')

    {

  
   
 return 1;

    }

    

    if(flag ==
1) // 1: left

    {

  
   
 for(int i = direction-1; i <=
direction+2; ++i)

  
   
 {

  
   
   
 newdir = (i+4)%4;

  
   
   
 int xx = sx + dir[newdir][0];