zoukankan      html  css  js  c++  java
  • POJ-2251 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!



    #include <iostream>
    #include <queue>
    using namespace std;
    
    char maze[31][31][31];
    int l, r, c;
    int sx, sy, sz, gx, gy, gz;
    int dx[6] = {0, 1, 0, -1, 0, 0};
    int dy[6] = {1, 0, -1, 0, 0, 0};
    int dz[6] = {0, 0, 0, 0, 1, -1};
    
    
    typedef struct P
    {
        int z;
        int x;
        int y;
        
    }P;
    
    int main(void)
    {
        int bfs(void);
        
        while(scanf("%d%d%d", &l, &r, &c) == 3 && l && r && c)
        {
            for(int i = 0; i < l; i++)
            {
                for(int j = 0; j < r; j++)
                {
                    scanf("%s", maze[i][j]);
                    for(int k = 0; k < c; k++)
                    {
                        if(maze[i][j][k] == 'S')
                        {
                            sx = j;
                            sy = k;
                            sz = i;
                        }
                        if(maze[i][j][k] == 'E')
                        {
                            gx = j;
                            gy = k;
                            gz = i;
                        }
                        
                    }
                }
            }
            
            int e = bfs();
            if(e == -1)
                printf("Trapped!
    ");
            else
                printf("Escaped in %d minute(s).
    ", e);
            
            
            
        }
        
        
        
        return 0;
    }
    
    
    
    int bfs(void)
    {
        queue<P> que;
        int data[31][31][31];
        P cas;
        
        memset(data, -1, sizeof(data));
        
        data[sz][sx][sy] = 0;
        cas.z = sz;
        cas.x = sx;
        cas.y = sy;
        que.push(cas);
        
        while(que.size())
        {
            P ca;
            ca = que.front();   que.pop();
            if(ca.z == gz && ca.x == gx && ca.y == gy)
                break;
            
            for(int i = 0; i < 6; i++)
            {
                cas.z = ca.z + dz[i];
                cas.x = ca.x + dx[i];
                cas.y = ca.y + dy[i];
                if(cas.z >= 0 && cas.z < l && cas.x >= 0 && cas.x < r && cas.y >= 0 && cas.y < c && maze[cas.z][cas.x][cas.y] != '#' && data[cas.z][cas.x][cas.y] < 0)
                {
                    que.push(cas);
                    data[cas.z][cas.x][cas.y] = data[ca.z][ca.x][ca.y] + 1;
                }
                
            }
            
            
        }
        
        
        return data[gz][gx][gy];
    }
    
  • 相关阅读:
    JS闭包
    js Date日期对象的扩展
    python通过post提交数据的方法
    python通过post提交数据的方法
    在Python中操作文件之truncate()方法的使用教程
    在Python中操作文件之truncate()方法的使用教程
    大数据将使安全产品爆发式增长
    大数据将使安全产品爆发式增长
    Python中内置数据类型list,tuple,dict,set的区别和用法
    Python中内置数据类型list,tuple,dict,set的区别和用法
  • 原文地址:https://www.cnblogs.com/limyel/p/7146010.html
Copyright © 2011-2022 走看看