• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>

            comiz

            a problem of maze

            Problem Statement

            People enjoy mazes, but they also get them dirty. Arrows, graffiti, and chewing gum are just a few of the souvenirs people leave on the walls. You, the maze keeper, are assigned to whiten the maze walls. Each face of the wall requires one liter of paint, but you are only required to paint visible faces. You are given a map of the maze, and you must determine the amount of paint needed for the job.

            The maze is described by a vector <string> maze, where each character can be either '#' (a wall) or '.' (an empty space). All '.' characters on the perimeter of the map are considered entrances to the maze. Upon entering the maze, one can only move horizontally and vertically through empty spaces, and areas that are not reachable by these movements are not considered visible. Each '#' represents a square block with four wall faces (each side of the square is a face). A face is visible if it is not directly adjacent to another wall (and is in a reachable area of the maze). For example, two adjacent blocks can have at most six visible faces since two of their faces are directly adjacent to each other. All exterior faces on the perimeter are considered visible.

            For example, the following picture represents a trivial maze with just one (wide) entrance and only four empty reachable spaces:

             TroytownKeeper.png

            To whiten this maze you must paint the faces highlighted in yellow above: 16 for its perimeter, plus 8 interior faces. Note that there are faces that are not visible and thus need not be painted.

            Definition     

            Class: TroytownKeeper

            Method: limeLiters Parameters: vector <string>

            Returns: int

            Method signature: int limeLiters(vector <string> maze)

            (be sure your method is public)     

            Constraints

            - maze will contain between 1 and 50 elements, inclusive.

            - Each element of maze will contain between 1 and 50 characters, inclusive.

            - All elements of maze will have the same number of characters.

            - All characters in maze will be either '.' or '#' . Examples 0)  

             

               

            {"##..#",
            "#.#.#",
            "#.#.#",
            "#####"}
            Returns: 24

            posted on 2007-11-04 19:35 comiz 閱讀(407) 評論(1)  編輯 收藏 引用

            評論

            # re: a problem of maze 2007-11-04 19:35 comiz

            using System;
            using System.Collections;
            using System.ComponentModel;
            using System.Data;
            using System.Threading;

            namespace TroytownKeeper
            {

            public class TroytownKeeper
            {
            string [] maze;
            bool [,]used=new bool[100,100];
            int sum=0;
            public TroytownKeeper()
            {



            }

            public int LimeLiters(string [] maze)
            {
            this.maze=maze;
            for(int x=0;x<maze.GetLength(0);x++)
            {
            if(maze[x][0]=='.') dfs(x,0);
            if(maze[x][maze[0].Length-1]=='.') dfs(x,maze[0].Length-1);
            }
            for(int y=0;y<maze[0].Length-1;y++)
            {
            if(maze[0][y]=='.') dfs(0,y);
            if(maze[maze.GetLength(0)-1][y]=='.') dfs(maze.GetLength(0)-1,y);
            }

            for(int x=0;x<maze.GetLength(0);x++)
            for(int y=0;y<maze[0].Length;y++)
            if(maze[x][y]=='#')
            {
            //upside
            if(x==0)
            sum++;
            if(x<maze.Length-1&&used[x+1,y])
            sum++;
            //leftside
            if(y==0)
            sum++;
            if(y>0&&used[x,y-1])
            sum++;
            //underside
            if(x==maze.Length-1)
            sum++;
            if(x>0&&used[x-1,y])
            sum++;
            //rightside
            if(y==maze[0].Length-1)
            sum++;
            if(y<maze[0].Length-1&&used[x,y+1])
            sum++;
            }
            return sum;
            }

            static void Main(string[] args)
            {
            TroytownKeeper TK=new TroytownKeeper();
            string [] str={"##..#"
            ,"#.#.#"
            ,"#.#.#"
            ,"#####"};
            int count=TK.LimeLiters(str);
            Console.WriteLine(count.ToString());
            }


            void dfs(int x,int y)
            {
            used[x,y]=true;
            if(x<maze.GetLength(0)&&maze[x+1][y]=='.'&&!used[x+1,y]) dfs(x+1,y);
            if(y<maze[0].Length&&maze[x][y+1]=='.'&&!used[x,y+1]) dfs(x,y+1);
            if(x>0&&maze[x-1][y]=='.'&&!used[x-1,y]) dfs(x-1,y);
            if(y>0&&maze[x][y-1]=='.'&&!used[x,y-1]) dfs(x,y-1);
            }
            }
            }

              回復  更多評論   

            <2025年6月>
            25262728293031
            1234567
            891011121314
            15161718192021
            22232425262728
            293012345

            導航

            統計

            常用鏈接

            留言簿(1)

            隨筆檔案

            搜索

            最新評論

            閱讀排行榜

            評論排行榜

            久久精品无码一区二区三区日韩 | 亚洲综合伊人久久大杳蕉| 久久青草国产手机看片福利盒子| 欧美日韩中文字幕久久伊人| 天天久久狠狠色综合| 久久久WWW成人免费精品| 中文字幕无码久久人妻| 久久91精品国产91久久户| 久久国产福利免费| 色欲久久久天天天综合网精品 | 无码久久精品国产亚洲Av影片| 国产成人精品久久一区二区三区 | 亚洲AV成人无码久久精品老人| 久久婷婷五月综合色奶水99啪| 东方aⅴ免费观看久久av| 久久成人国产精品二三区| 亚洲国产成人久久综合一区77| 亚洲国产欧洲综合997久久| 国产综合免费精品久久久| 久久久SS麻豆欧美国产日韩| 99久久精品免费看国产免费| 国产色综合久久无码有码| 99精品伊人久久久大香线蕉| 亚洲精品无码久久久久去q | 无码人妻精品一区二区三区久久久 | 免费观看久久精彩视频| 久久精品国产亚洲沈樵| 久久99精品国产麻豆宅宅| 无码国内精品久久综合88 | 91精品婷婷国产综合久久| 久久久久亚洲av成人网人人软件| 狠狠色丁香久久综合五月| 久久久久人妻一区精品性色av| 久久精品亚洲精品国产欧美| 精品久久久久久久| 久久精品无码专区免费青青| 亚洲伊人久久成综合人影院| 久久久久久国产精品免费免费| 久久精品人人做人人爽电影| 无码超乳爆乳中文字幕久久 | 久久久噜噜噜久久中文字幕色伊伊|