zoukankan      html  css  js  c++  java
  • POJ 2251, Dungeon Master

    Time Limit: 1000MS  Memory Limit: 65536K
    Total Submissions: 5685  Accepted: 2256


    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


    // POJ2251.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    #include 
    <queue>
    using namespace std;
    struct Point
    {
        
    int x,y,z;
    };
    int main(int argc, char* argv[])
    {
        
    char c[31][31][31];
        
    int cube[31][31][31];
        
    int L,R,C;
        
    int step[6][3= {-1,0,01,0,00,-1,00,1,00,0,-10,0,1};

        
    while(scanf("%d %d %d\n",&L, &R, &C) && L!= 0 && R!= 0 && C!= 0)
        {
            Point pt;
            
    for (int i = 0; i < L; ++i)
                
    for (int j = 0; j <= R; ++j)
                    gets(c[i][j]);

            
    for (int i = 0; i < L; ++i)
                
    for (int j = 0; j < R; ++j)
                    
    for (int k = 0; k < C; ++k)
                    {
                        
    if (c[i][j][k] == 'S')
                        {
                            pt.x 
    = k;
                            pt.y 
    = j;
                            pt.z 
    = i;
                            cube[i][j][k] 
    = 0;
                        }
                        
    else if (c[i][j][k] == '.') cube[i][j][k] = 0;
                        
    else if (c[i][j][k] == '#') cube[i][j][k] = -1;
                        
    else if (c[i][j][k] == 'E') cube[i][j][k] = -2;
                    };

            queue
    <Point> q;
            q.push(pt);

            
    int cnt = -1;
            
    while (!q.empty())
            {
                Point p 
    = q.front();
                q.pop();
                
    for (int i = 0; i < 6++i)
                {
                    Point next;
                    next.x 
    = p.x + step[i][0];
                    next.y 
    = p.y + step[i][1];
                    next.z 
    = p.z + step[i][2];
                    
    if (next.x >= 0 && next.x < C && next.y >= 0 && next.y < R && next.z >= 0 && next.z < L)
                    {
                        
    if (cube[next.z][next.y][next.x] == 0)
                        {
                            cube[next.z][next.y][next.x] 
    = cube[p.z][p.y][p.x] + 1;
                            q.push(next);
                        }
                        
    else if (cube[next.z][next.y][next.x] == -2)
                        {
                            cnt 
    = cube[p.z][p.y][p.x] + 1;
                            
    break;
                        }
                    }
                }

                
    if (cnt != -1break;
            };

            
    if (cnt == -1) cout <<"Trapped!\n";
            
    else cout << "Escaped in "<<cnt<<" minute(s).\n";
        };
        
    return 0;
    }
  • 相关阅读:
    全国最全的省,市,县,电话号前缀,邮编数据
    数组的高级用法
    Maven Project configuration is not up-to-date with pom.xml错误解决方法
    HighCharts开发说明
    Java基础复习之二:运算符,键盘录入,流程控制语句,if语句,三元运算
    Java基础复习之一篇:关健字,标识符,注释,常量,进制转换,变量,数据类型,数据类型转换
    ehcache 缓存技术
    Write operations are not allowed in read-only mode
    浮点运算与boost.multiprecision
    关于OpenCASCADE数组序列的起始值
  • 原文地址:https://www.cnblogs.com/asuran/p/1580199.html
Copyright © 2011-2022 走看看