zoukankan      html  css  js  c++  java
  • Dungeon Master

    Dungeon Master
    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

    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!




    最简单BFS,直接代码:

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    #include<vector>
    #include<map>
    #include<queue>
    #include<stack>
    #include<string>
    #include<map>
    #include<set>
    #include<ctime>
    #define eps 1e-6
    #define MAX 100005
    #define INF 0x3f3f3f3f
    #define LL long long
    #define pii pair<int,int>
    #define rd(x) scanf("%d",&x)
    #define rd2(x,y) scanf("%d%d",&x,&y)
    #define rd3(x,y,z) scanf("%d%d%d",&x,&y,&z)
    ///map<int,int>mmap;
    ///map<int,int >::iterator it;
    using namespace std;
    int dir[6][3]={{-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,1},{0,0,-1}};
    int level,row,col;
    char mmap[35][35][35];
    bool vis[35][35][35];
    
    struct Pos
    {
        int l;
        int r;
        int c;
        int coun;
        Pos(){}
        Pos(int l,int r,int c,int coun)
        {
            this->l=l,this->r=r,this->c=c,this->coun = coun;
        }
    }start,eend;
    
    bool Ok(int l,int r,int c){
      return (mmap[l][r][c]=='.'  && l>=0 && r>=0 && c>=0 && l<level && r<row && c<col && vis[l][r][c]==false);
    }
    
    int main ()
    {
        while(~scanf("%d%d%d",&level,&row,&col)&&level!=0&&row!=0&&col!=0)
        {
            memset(vis,0,sizeof(vis));
            for(int i=0; i<level; i++)
                for(int j=0; j<row; j++)
                {
                    scanf("%s",mmap[i][j]);
                    for(int k=0; k<col; k++)
                    {
                        if(mmap[i][j][k] == 'S')
                          start.l=i,start.r = j,start.c = k,start.coun=0;
                        if(mmap[i][j][k] == 'E')
                         eend.l=i,eend.r = j,eend.c = k,mmap[i][j][k] = '.';
                    }
                }
            queue<Pos> que;
            que.push(start);
            vis[start.l][start.r-1][start.c]=true;
            int coun = -1,come = 0;
            while(!que.empty())
            {
                Pos a=que.front();
                que.pop();
                if(a.l==eend.l&&a.r==eend.r&&a.c==eend.c)
                {
                    come = 1;
                    coun = a.coun;
                    break;
                }
                for(int i=0;i<6;i++){
                    int l=a.l+dir[i][0],r=a.r+dir[i][1],c=a.c+dir[i][2];
                    if( Ok(l,r,c)  )
                    {
                        que.push(Pos(l,r,c,a.coun+1));
                        vis[l][r][c]=1;
                    }
                }
            }
            if(come)  printf("Escaped in %d minute(s).
    ",coun);
            else  printf("Trapped!
    ");
        }
        return 0;
    }
    




  • 相关阅读:
    0302思考并回答一些问题
    1231 实验四 递归下降语法分析程序设计
    1118 实验三 有限自动机的构造与识别
    1112对他人的博客评论及建议
    1029c语言文法
    1022词法分析实验总结
    词法分析
    0330 复利计算——单元测试
    0321 读《构建之法》第一,二,三章有感
    0316 复利计算总结(0330 更新)
  • 原文地址:https://www.cnblogs.com/zswbky/p/6717991.html
Copyright © 2011-2022 走看看