zoukankan      html  css  js  c++  java
  • virtual judge(专题一 简单搜索 B)

    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!

    三维迷宫水题
    #include"cstdio"
    #include"queue"
    #include"cstring"
    using namespace std;
    const int MAXN=31;
    char dun[MAXN][MAXN][MAXN];
    int vis[MAXN][MAXN][MAXN];
    struct node{
        int L,R,C;
        int step;
        node(int l,int r,int c,int s):L(l),R(r),C(c),step(s){}
        node(){}
    };
    int l,r,c;
    int sl,sr,sc;
    int el,er,ec;
    int dl[6]={1,0,0,0,0,-1};
    int dc[6]={0,1,0,-1,0,0};
    int dr[6]={0,0,1,0,-1,0};
    int bfs()
    {
        queue<node> que;
        que.push(node(sl,sr,sc,0));
        vis[sl][sr][sc]=1;
        while(!que.empty())
        {
            node now=que.front();que.pop();
            if(now.L==el&&now.R==er&&now.C==ec)
            {
                return now.step;
            }
            for(int i=0;i<6;i++)
            {
                int nl=now.L+dl[i];
                int nr=now.R+dr[i];
                int nc=now.C+dc[i];
                if(0<=nl&&nl<l&&0<=nr&&nr<r&&0<=nc&&nc<c&&!vis[nl][nr][nc]&&dun[nl][nr][nc]!='#')
                {
                    vis[nl][nr][nc]=1;
                    que.push(node(nl,nr,nc,now.step+1));
                }
            }
        }
        return -1;
    }
    int main()
    {
        while(scanf("%d%d%d",&l,&r,&c)!=EOF&&l!=0)
        {
            memset(vis,0,sizeof(vis));
            scanf("%*c");
            for(int i=0;i<l;i++)
            {
                for(int j=0;j<r;j++)
                {
                    for(int z=0;z<c;z++)
                    {
                        scanf("%c",&dun[i][j][z]);
                        if(dun[i][j][z]=='S')
                        {
                            sl=i,sr=j,sc=z;
                        }
                        if(dun[i][j][z]=='E')
                        {
                            el=i,er=j,ec=z;
                        }
                    }
                    scanf("%*c");
                }
                scanf("%*c");
            }
            
        int ans=bfs();
        if(ans!=-1)
        {
            printf("Escaped in %d minute(s).
    ",ans);
        }
        else
        {
            printf("Trapped!
    ");
        }
    }
        return 0;
    }
  • 相关阅读:
    JAVA基础学习之路(九)[2]String类常用方法
    [MYSQL]练习(一)
    JAVA基础学习之路(十一)引用传递
    java--多线程编程简介
    序列化和反序列化的理解
    简单的socket编程
    php-生成数据库设计文档
    centos7 jenkins安装和使用
    centos7 rabbitmq安装以及应用
    centos7 dubbokeeper安装
  • 原文地址:https://www.cnblogs.com/program-ccc/p/5001706.html
Copyright © 2011-2022 走看看