POJ2251 Dungeon Master 解题报告

分类：BFS，迷宫，STL，队列

作者：ACShiryu

时间：2011-7-23

地址：http://www.cnblogs.com/ACShiryu/archive/2011/07/23/2114994.html

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9395		Accepted: 3652

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!

题目大意：这题是一个三维的迷宫题目，其中用'.'表示空地，'#'表示障碍物，'S'表示起点，'E'表示终点，求从起点到终点的最小移动次数，解法和二维的类似，只是在行动时除了东南西北移动外还多了上下。

对于题目给出数据的含义就是输入l，r，c，分别代表迷宫有l层，每层长宽分别是c，r。

对于数据以可以这样移动

(1,1,1)->(1,1,2)->(1,1,3)->(1,1,4)->(1,1,5)->(1,2,5)

->(1,3,5)->(1,3,4)->(1,4,4)->(2,4,4)->(2,4,5)->(3,4,,5)

共11步就可以到达终点 

对于数据二明显不能到达，则输出Trapped

这题用BFS解，每次去队首元素，如果是终点则输出结果移动的次数，否则，从该点开始分别向东南西北上下移动（如果可以走的话）并继续搜，如果到队列为空还没搜到解法，则说明无解。

这题我提交一次就AC了，可以看出难度不是很大。

参考代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
bool hash[35][35][35];
char map[35][35][35];
struct prog{
    int x;int y;int z;        //定义点的坐标
    int step;                //移动的步子数
};
int main()
{
    int l , r , c ;
    while ( cin >> l >> r >> c , l && r && c )
    {
        int i , j , k;
        prog start;
        memset(hash,false,sizeof(hash));//初始化为false，表示每条路都没走过
        for ( i = 0 ; i < l ; i ++ )
        {
            for ( j = 0 ; j < r ; j ++ )
            {
                cin >> map [i][j];
                for ( k = 0 ; k < c ; k ++ )
                {
                    if(map[i][j][k]=='S')
                    {//对起点的相关变量初始化
                        start.x=i;
                        start.y=j;
                        start.z=k;
                        start.step=0;
                        hash[i][j][k]=true;
                    }
                }
            }
        }

        queue<prog>bfs;
        bfs.push(start);    //起点入队
        bool found=false;    //标记变量，判断是否找到最优解，如果为true则说明从起点到终点存在着解
        
        while(!bfs.empty())
        {
            prog tmp=bfs.front();
            bfs.pop();

            if(map[tmp.x][tmp.y][tmp.z]=='E')
            {//已经到了终点，则将标记变量设为true，并输出答案
                found=true;
                cout<<"Escaped in "<<tmp.step<<" minute(s)."<<endl;
                break;
            }
            
            tmp.step++;        //移动步子+1
            prog tmp2;
            if(tmp.x>0)
            {//向下走，所以改点不能在第一层，则x要大于0
                tmp2=tmp;
                tmp2.x--;
                if(map[tmp2.x][tmp2.y][tmp2.z]!='#'&&hash[tmp2.x][tmp2.y][tmp2.z]==false)
                {//如果下一层不是障碍物并且也没走过，则往下走，并标记该点已走过，入队
                    hash[tmp2.x][tmp2.y][tmp2.z]=true;
                    bfs.push(tmp2);
                }
            }

            //下面搜索方法同上 注释略
            if(tmp.x<l-1)
            {//向上
                tmp2=tmp;
                tmp2.x++;
                if(map[tmp2.x][tmp2.y][tmp2.z]!='#'&&hash[tmp2.x][tmp2.y][tmp2.z]==false)
                {
                    hash[tmp2.x][tmp2.y][tmp2.z]=true;
                    bfs.push(tmp2);
                }
            }

            if(tmp.y>0)
            {//向北
                tmp2=tmp;
                tmp2.y--;
                if(map[tmp2.x][tmp2.y][tmp2.z]!='#'&&hash[tmp2.x][tmp2.y][tmp2.z]==false)
                {
                    hash[tmp2.x][tmp2.y][tmp2.z]=true;
                    bfs.push(tmp2);
                }
            }

            if(tmp.y<r-1)
            {//向南
                tmp2=tmp;
                tmp2.y++;
                if(map[tmp2.x][tmp2.y][tmp2.z]!='#'&&hash[tmp2.x][tmp2.y][tmp2.z]==false)
                {
                    hash[tmp2.x][tmp2.y][tmp2.z]=true;
                    bfs.push(tmp2);
                }
            }

            if(tmp.z>0)
            {//向西
                tmp2=tmp;
                tmp2.z--;
                if(map[tmp2.x][tmp2.y][tmp2.z]!='#'&&hash[tmp2.x][tmp2.y][tmp2.z]==false)
                {
                    hash[tmp2.x][tmp2.y][tmp2.z]=true;
                    bfs.push(tmp2);
                }
            }

            if(tmp.z<c-1)
            {//向东
                tmp2=tmp;
                tmp2.z++;
                if(map[tmp2.x][tmp2.y][tmp2.z]!='#'&&hash[tmp2.x][tmp2.y][tmp2.z]==false)
                {
                    hash[tmp2.x][tmp2.y][tmp2.z]=true;
                    bfs.push(tmp2);
                }
            }

        }
        if(!found)//如果没有找到解
            cout<<"Trapped!"<<endl;
    }
    return 0;
}