POJ2251Dungeon Master（图搜）（三维bfs）

本文介绍了一种解决三维迷宫逃脱问题的算法，通过广度优先搜索（BFS）来寻找从起点到终点的最短路径。输入包括迷宫的尺寸、层数、行数、列数以及迷宫的地图描述，输出则是逃脱所需的时间或无法逃脱的提示。

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Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22503		Accepted: 8787

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

不用剪枝的图搜，思路还是比较简单的，只不过从二维图变成了三维图，这里初始化都要用三维的就行了，二维的走法是4种，三维的走法是六种，上下左右前后，思路很好写，直接就能1A 注意数组不要开太大，POJ的范围给的还是蛮准的。

这里直接上AC代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
struct zuobiao
{
    int x,y,z;
}now,nex;
int q,n,m;
char a[32][32][32];
int vis[32][32][32];
int output[32][32][32];
int fx[6] = {0, 0, 0, 0, 1, -1};
int fy[6] = {0, 0, 1, -1, 0, 0};
int fz[6] = {1, -1, 0, 0, 0, 0};
void bfs(int x,int y,int z)
{
    memset(vis,0,sizeof(vis));
    memset(output,0,sizeof(output));
    vis[x][y][z]=1;
    now.x=x;
    now.y=y;
    now.z=z;
    queue<zuobiao >s;
    s.push(now);
    while(!s.empty())
    {
        now=s.front();
        if(a[now.x][now.y][now.z]=='E')
        {
            //printf("%d\n",output[nex.x][nex.y][nex.z]);
            return ;
        }
        s.pop();
        for(int i=0;i<6;i++)
        {
            nex.x=now.x+fx[i];
            nex.y=now.y+fy[i];
            nex.z=now.z+fz[i];
            if(nex.x>=0&&nex.x<q&&nex.y>=0&&nex.y<n&&nex.z>=0&&nex.z<m&&vis[nex.x][nex.y][nex.z]==0&&a[nex.x][nex.y][nex.z]!='#')
            {
                output[nex.x][nex.y][nex.z]=output[now.x][now.y][now.z]+1;
                vis[nex.x][nex.y][nex.z]=1;
                s.push(nex);
            }
        }
    }
    return ;
    //printf("no\n");
}
int main()
{
    while(~scanf("%d%d%d",&q,&n,&m))
    {
        if(q==0&&n==0&&m==0)break;
        int x,y,z;
        int ex,ey,ez;
        for(int i=0;i<q;i++)
        {
            for(int j=0;j<n;j++)
            {
                scanf("%s",a[i][j]);
                for(int k=0;k<m;k++)
                {
                    if(a[i][j][k]=='S')
                    {
                        x=i;
                        y=j;
                        z=k;
                    }
                    if(a[i][j][k]=='E')
                    {
                        ex=i;
                        ey=j;
                        ez=k;
                    }
                }
            }
        }
        bfs(x,y,z);
        if(output[ex][ey][ez]==0){printf("Trapped!\n");continue;}
        printf("Escaped in %d minute(s).\n",output[ex][ey][ez]);
    }
}