• <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(407) | 評(píng)論 (1)編輯 收藏

            2007年10月24日

            一道基礎(chǔ)題

            12,…,99個(gè)數(shù)分成三組,分別組成三個(gè)三位數(shù),且使這三個(gè)三位數(shù)構(gòu)成

               123的比例,試求出所有滿足條件的三個(gè)三位數(shù)。

               例如:三個(gè)三位數(shù)192,384,576滿足以上條件。
            題目比較基礎(chǔ),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(212) | 評(píng)論 (0)編輯 收藏

            僅列出標(biāo)題  
            <2025年6月>
            25262728293031
            1234567
            891011121314
            15161718192021
            22232425262728
            293012345

            導(dǎo)航

            統(tǒng)計(jì)

            常用鏈接

            留言簿(1)

            隨筆檔案

            搜索

            最新評(píng)論

            閱讀排行榜

            評(píng)論排行榜

            中文字幕乱码久久午夜| 亚洲欧美伊人久久综合一区二区| 人妻精品久久无码专区精东影业| 99久久无色码中文字幕人妻| 国产精品99久久久精品无码| 国产美女亚洲精品久久久综合| 亚洲中文字幕久久精品无码喷水 | 色综合久久久久无码专区 | 久久天堂电影网| 97精品伊人久久久大香线蕉 | 欧美日韩成人精品久久久免费看| 午夜福利91久久福利| 婷婷久久香蕉五月综合加勒比 | 国产V综合V亚洲欧美久久| 久久久久无码精品| 久久亚洲精精品中文字幕| 亚洲国产精品久久久久网站 | 99国内精品久久久久久久| 伊人久久无码精品中文字幕| 久久夜色精品国产欧美乱| 久久99精品久久久久久水蜜桃| 无码人妻精品一区二区三区久久 | 免费精品99久久国产综合精品| 免费一级做a爰片久久毛片潮| 亚洲第一极品精品无码久久| 久久九九久精品国产| 国内精品久久久久伊人av| 久久只有这里有精品4| 国产精品久久久久久一区二区三区| 久久国产高清一区二区三区| 精品久久久久久久久午夜福利| 国产欧美久久久精品影院| 久久久久国色AV免费观看| 精品久久香蕉国产线看观看亚洲| 久久这里的只有是精品23| 国产免费福利体检区久久| 国产成人精品久久免费动漫| 亚洲精品第一综合99久久| 久久久中文字幕日本| 久久精品国产精品青草| 久久精品国产亚洲av麻豆色欲|