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            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(212) | 評論 (0)編輯 收藏

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