zoukankan      html  css  js  c++  java
  • POJ 2251 Dungeon Master(bfs)

    Dungeon Master

     

    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!

     

     1 #include<cstdio>
     2 #include<queue>
     3 #include<cstring>
     4 using namespace std;
     5 
     6 struct point
     7 {
     8     int x,y,z;
     9     int step;
    10 };
    11 
    12 int sx,sy,sz,ex,ey,ez;
    13 int L,R,C;
    14 char map[32][32][32];
    15 int vis[32][32][32];
    16 int d[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,0,1},{0,-1,0},{0,0,-1}};
    17 
    18 bool check(point node)
    19 {
    20     if(node.x==ex&&node.y==ey&&node.z==ez)
    21     return true;
    22     return false;
    23 }
    24 
    25 int bfs(int z,int x,int y)
    26 {
    27     queue<point>q;
    28     point init;
    29     init.x=x,init.y=y,init.z=z;
    30     init.step=0;
    31     q.push(init);
    32     vis[z][x][y]=1;
    33     while(!q.empty())
    34     {
    35         point t1,t2;
    36         t1=q.front();
    37         q.pop();
    38         if(check(t1))
    39         return t1.step;
    40         for(int i=0;i<6;i++)
    41         {
    42             t2.z=t1.z+d[i][0];
    43             t2.x=t1.x+d[i][1];
    44             t2.y=t1.y+d[i][2];
    45             t2.step=t1.step+1;
    46             if(!vis[t2.z][t2.x][t2.y]&&map[t2.z][t2.x][t2.y]!='#')
    47             {
    48                 q.push(t2);
    49                 vis[t2.z][t2.x][t2.y]=1;
    50             }
    51         }
    52     }
    53     return -1;
    54 }
    55 
    56 int main()
    57 {
    58     //freopen("in.txt","r",stdin);
    59     char temp;
    60     int i,j,k;
    61     while(scanf("%d%d%d",&L,&R,&C),L||R||C)
    62     {
    63         memset(map,'#',sizeof(map));
    64         memset(vis,0,sizeof(vis));
    65         for(i=1;i<=L;i++)
    66         {
    67             for(j=1;j<=R;j++)
    68             {
    69                 getchar();
    70                 for(k=1;k<=C;k++)
    71                 {
    72                     scanf("%c",&temp);
    73                     map[i][j][k]=temp;
    74                     if(temp=='S')
    75                     sz=i,sx=j,sy=k;
    76                     else if(temp=='E')
    77                     ez=i,ex=j,ey=k;
    78                     else;
    79                 }
    80             }
    81             getchar();
    82         }
    83         int ans=bfs(sz,sx,sy);
    84         if(ans==-1)
    85         printf("Trapped!
    ");
    86         else
    87         printf("Escaped in %d minute(s).
    ",ans);
    88     }
    89     return 0;
    90 }
  • 相关阅读:
    为什么不能直接导入Statsmodels使用?
    数据分析工作的主要内容和基本流程
    Nodejs 包与 NPM 第三方模块安装和 package.json 以及 CNPM (4)
    CommonJs 和 Nodejs 中自定义模块 (3)
    pyhthon 处理pdf 合集
    02 nodejs HTTP模块和url模块配置supervisor
    1 nodejs简介与开发环境配置
    mysql 修改root密码和禁用无密码登录配置
    floodFill填充函数函数(六)
    粗略的调整图片对比度和亮度(五)
  • 原文地址:https://www.cnblogs.com/homura/p/4705890.html
Copyright © 2011-2022 走看看