• <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

            2007年11月4日

            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 @ 2007-11-04 19:35 comiz 閱讀(409) | 評論 (1)編輯 收藏

            2007年10月24日

            一道基礎題

            12,…,99個數分成三組,分別組成三個三位數,且使這三個三位數構成

               123的比例,試求出所有滿足條件的三個三位數。

               例如:三個三位數192,384,576滿足以上條件。
            題目比較基礎,自己用的回朔法,萬里高樓平地起,慢慢來吧...
            /* Note:Your choice is C IDE */
            #define null 0
            #include "stdio.h"
            void inject(int N,int *nNum)
            {
                int sum[3],i,j,k;
                if(N==0)
                {
                    sum[0]=*nNum*100+*(nNum+1)*10+*(nNum+2);
                    sum[1]=*(nNum+3)*100+*(nNum+4)*10+*(nNum+5);
                    sum[2]=*(nNum+6)*100+*(nNum+7)*10+*(nNum+8);
                    if(((sum[0]<<1)==sum[1])&&((3*sum[0])==sum[2]))
                    {
                        printf("we have one of them:");   
                        printf("%d,%d,%d\n",sum[0],sum[1],sum[2]);
                    }
                }
                else
                {
                    for(j=0;j<9;j++)
                    {
                        if(*(nNum+j)==null)
                        {
                            *(nNum+j)=N;
                            inject(N-1,nNum);
                            *(nNum+j)=null;
                        }
                    }
                }
            }
            main()
            {
                int k;
                int Num[9];
                for(k=0;k<9;k++)
                {
                    Num[k]=null;   
                }
                    inject(9,Num);
            }

            posted @ 2007-10-24 18:22 comiz 閱讀(213) | 評論 (0)編輯 收藏

            僅列出標題  
            <2025年7月>
            293012345
            6789101112
            13141516171819
            20212223242526
            272829303112
            3456789

            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

            青青青青久久精品国产h久久精品五福影院1421 | 亚洲精品97久久中文字幕无码| 久久精品男人影院| 久久亚洲AV永久无码精品| 亚洲中文久久精品无码ww16| 99久久久国产精品免费无卡顿| 亚洲国产精品久久久久| 久久久久久久久久久精品尤物| 久久久久久久97| 日韩久久久久中文字幕人妻| 久久久久久无码Av成人影院| 午夜精品久久久内射近拍高清| 99re这里只有精品热久久| 欧洲国产伦久久久久久久 | 久久精品国产半推半就| 国产欧美久久久精品影院| 一级做a爰片久久毛片人呢| 亚洲午夜久久久久久久久久| 久久无码精品一区二区三区| 成人久久精品一区二区三区| 久久人妻少妇嫩草AV无码专区| 久久强奷乱码老熟女网站| 久久精品九九亚洲精品天堂 | 无码日韩人妻精品久久蜜桃| 国产精品热久久无码av| 久久精品国产福利国产秒| 久久99精品国产自在现线小黄鸭 | 久久国产欧美日韩精品| 久久婷婷五月综合成人D啪| 精品熟女少妇aⅴ免费久久| 亚洲成色999久久网站| 狠狠久久亚洲欧美专区| 精品永久久福利一区二区| 精品国际久久久久999波多野| 亚洲精品乱码久久久久久中文字幕 | 色欲久久久天天天综合网 | 亚洲精品乱码久久久久久蜜桃图片 | 久久精品国内一区二区三区| 久久婷婷久久一区二区三区| 精品国产91久久久久久久| 久久天堂电影网|