zoj 1940 Dungeon Master

B - Dungeon Master
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld
& %llu
Submit Status Practice ZOJ
1940

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!

2014年第一道搜索题目，出了很多问题，这是一道3维基础广搜模板题目，记录一下。

#include <cstdio>
#include<iostream>
#include <queue>
using namespace std;
const int N = 85;
char map

;   //zhuyi
int l,r,c;
struct Node
{
    int x,y,z;
    int step;
};

int dir[8][4]=
{
    {0,0,1},
    {0,0,-1},
    {0,1,0},
    {0,-1,0},
    {1,0,0},
    {-1,0,0}
};

int Bfs(Node st,Node en)
{
    queue<Node> q;
    Node tmd,tmp;
    q.push(st);
    while(!q.empty())
    {
        tmd=q.front();q.pop();   //初始片段搜索结束条件的地方
        if(tmd.x==en.x && tmd.y==en.y && tmd.z==en.z)
            return tmd.step;
        for(int i=0;i<6;i++)
        {
            tmp.x=tmd.x+dir[i][0];
            tmp.y=tmd.y+dir[i][1];
            tmp.z=tmd.z+dir[i][2];
            tmp.step=tmd.step+1;
            if(map[tmp.x][tmp.y][tmp.z]!='#'&&tmp.x>=0 &&tmp.y>=0 && tmp.z>=0 &&tmp.x<l&&tmp.y<r&&tmp.z<c)
            { //注意上面的搜索条件，是！=‘#’ 错在了这里
                q.push(tmp);
                map[tmp.x][tmp.y][tmp.z]='#';

            }
        }
    }
    return -1;
}
int main()
{
    while(~scanf("%d%d%d",&l,&r,&c))
    {
        if(l==0 && r==0 &&c==0)
            break;
        getchar();
        Node st,en;
        for(int i=0;i<l;i++)
        {
            for(int j=0;j<r;j++)
            {
                for(int k=0;k<c;k++)
                {
                    scanf("%c",&map[i][j][k]);
                    if(map[i][j][k]=='S')
                    {
                        st.x=i; st.y=j; st.z=k;st.step=0;
                    }
                    if(map[i][j][k]=='E')
                    {
                        en.x=i,en.y=j,en.z=k;
                    }
                }
                getchar();
            }
            if(i!=l-1)
                getchar();
        }
       // cout<<st.x<<st.y<<st.z<<en.x<<en.y<<en.z<<endl;
        int ans = Bfs(st,en);
        if(ans == -1){
            printf("Trapped!\n");
        }else
        {
            printf("Escaped in %d minute(s).\n",ans);
        }/*
        for(int i=0;i<l;i++)
        {
            for(int j=0;j<r;j++)
            {
                for(int k=0;k<c;k++)
                    printf("%c ",map[i][j][k]);
                putchar('\n');
            }
            putchar('\n');
        }*/
    }
    return 0;
}