POJ 2251, Dungeon Master-优快云博客

本文详细解析了一种迷宫逃脱算法，通过三维迷宫模型，使用广度优先搜索策略寻找从起点到终点的最短路径。算法首先将起点标记为0，然后遍历其周围的单元格，将未被访问过的单元格标记为当前单元格步数加一，并将其加入队列中。此过程重复直到找到终点或队列为空。如果找到终点，则输出逃脱所需的时间；如果队列为空且未找到终点，则输出无法逃脱。

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Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5685 Accepted: 2256

Description
You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.

Is an escape possible? If yes, how long will it take?

Input
The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size).
L is the number of levels making up the dungeon.
R and C are the number of rows and columns making up the plan of each level.
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output
Each maze generates one line of output. If it is possible to reach the exit, print a line of the form
Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape.
If it is not possible to escape, print the line
Trapped!

Sample Input
3 4 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

Sample Output
Escaped in 11 minute(s).
Trapped!

Source
Ulm Local 1997

// POJ2251.cpp : Defines the entry point for the console application.
//

#include < iostream >
#include < queue >
using namespace std;
struct Point
{
     int x,y,z;
};
int main( int argc, char * argv[])
{
     char c[ 31 ][ 31 ][ 31 ];
     int cube[ 31 ][ 31 ][ 31 ];
     int L,R,C;
     int step[ 6 ][ 3 ] = { - 1 , 0 , 0 , 1 , 0 , 0 , 0 , - 1 , 0 , 0 , 1 , 0 , 0 , 0 , - 1 , 0 , 0 , 1 };

     while (scanf( " %d %d %d\n " , & L, & R, & C) && L != 0 && R != 0 && C != 0 )
    {
        Point pt;
         for ( int i = 0 ; i < L; ++ i)
             for ( int j = 0 ; j <= R; ++ j)
                gets(c[i][j]);

         for ( int i = 0 ; i < L; ++ i)
             for ( int j = 0 ; j < R; ++ j)
                 for ( int k = 0 ; k < C; ++ k)
                {
                     if (c[i][j][k] == ' S ' )
                    {
                        pt.x = k;
                        pt.y = j;
                        pt.z = i;
                        cube[i][j][k] = 0 ;
                    }
                     else if (c[i][j][k] == ' . ' ) cube[i][j][k] = 0 ;
                     else if (c[i][j][k] == ' # ' ) cube[i][j][k] = - 1 ;
                     else if (c[i][j][k] == ' E ' ) cube[i][j][k] = - 2 ;
                };

        queue < Point > q;
        q.push(pt);

         int cnt = - 1 ;
         while ( ! q.empty())
        {
            Point p = q.front();
            q.pop();
             for ( int i = 0 ; i < 6 ; ++ i)
            {
                Point next;
                next.x = p.x + step[i][ 0 ];
                next.y = p.y + step[i][ 1 ];
                next.z = p.z + step[i][ 2 ];
                 if (next.x >= 0 && next.x < C && next.y >= 0 && next.y < R && next.z >= 0 && next.z < L)
                {
                     if (cube[next.z][next.y][next.x] == 0 )
                    {
                        cube[next.z][next.y][next.x] = cube[p.z][p.y][p.x] + 1 ;
                        q.push(next);
                    }
                     else if (cube[next.z][next.y][next.x] == - 2 )
                    {
                        cnt = cube[p.z][p.y][p.x] + 1 ;
                         break ;
                    }
                }
            }

             if (cnt != - 1 ) break ;
        };

         if (cnt == - 1 ) cout << " Trapped!\n " ;
         else cout << " Escaped in " << cnt << " minute(s).\n " ;
    };
     return 0 ;
}

转载于:https://www.cnblogs.com/asuran/archive/2009/10/10/1580199.html