zoukankan      html  css  js  c++  java
  • poj2251_kuagnbin带你飞专题一

    Dungeon Master
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 32684   Accepted: 12529

    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

    解题思路:只是把二维的变成三维的了,本质上不变,就增加了两个方向。
    但我还是卡了好久,我刚开始只记录了上一步的位置,不往上一步的位置走,结果一直超时。
    后来发现只要走过的地方就把vis数组置为1,以后就不能走到这个位置了。
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    
    using namespace std;
    
    struct node{
        int x,y,z;
        int step;
    };
    
    int l,r,c;
    char maze[40][40][40];
    int vis[40][40][40];
    node start;
    node endd;
    int directx[7]={0,0,0,0,1,-1};
    int directy[7]={-1,1,0,0,0,0};
    int directz[7]={0,0,-1,1,0,0};
    
    int findd(){
        queue<node> que;
        memset(vis,0,sizeof(vis));
        start.step=0;
        que.push(start);
        while(!que.empty()){
            node now=que.front();
            que.pop();
            for(int i=0;i<6;i++){
                node next=now;
                next.x+=directx[i];
                next.y+=directy[i];
                next.z+=directz[i];
                int xx=next.x;
                int yy=next.y;
                int zz=next.z;
                if(maze[xx][yy][zz]=='#'||vis[xx][yy][zz]||!(xx>=0&&xx<l&&yy>=0&&yy<r&&yy>=0&&zz>=0&&zz<c)){
                    continue;
                }
                if(maze[xx][yy][zz]=='E'){
                    next.step+=1;
                    return next.step;
                }
                if(maze[xx][yy][zz]=='.'){
                    next.step+=1;
                    vis[xx][yy][zz]=1;
                    que.push(next);
                }
            }
        }
        return 0;
    }
    
    int main()
    {
        while(scanf("%d %d %d",&l,&r,&c)&&(l+r+c!=0)){
            getchar();
            for(int i=0;i<l;i++,getchar()){
                for(int j=0;j<r;j++,getchar()){
                    for(int k=0;k<c;k++){
                        scanf("%c",&maze[i][j][k]);
                        if(maze[i][j][k]=='S'){
                            start.x=i;
                            start.y=j;
                            start.z=k;
                        }
                    }
                }
            }
            int ans=findd();
            if(ans){
                printf("Escaped in %d minute(s).
    ",ans);
            }else{
                printf("Trapped!
    ");
            }
    
        }
    
    
        return 0;
    }
  • 相关阅读:
    浅谈值对象
    循环一个节点列表(NodeList)或者数组,并且绑定事件处理函数引发对闭包的理解
    当前窗口和Iframe之间的相互访问(图片上传成功后立刻显示在当前页面上)
    网页动态加载图片 通过JS和jquery实现。
    javascript拖动层函数封装
    javascript中变量声明提升(Hoisting)
    运动框架必备的运动算法 留着用!
    CSS3特性之改变在浏览器上选中文字时,默认的背景颜色和文字颜色
    仿淘宝商品图片放大镜效果(鼠标移动上去会出现放大的图片,并且可以移动)
    与PHP交互中文编码的几个函数 decodeURIComponent,encodeURIComponent,encodeURI,decodeURI
  • 原文地址:https://www.cnblogs.com/TWS-YIFEI/p/6696677.html
Copyright © 2011-2022 走看看