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            The Fourth Dimension Space

            枯葉北風(fēng)寒,忽然年以殘,念往昔,語默心酸。二十光陰無一物,韶光賤,寐難安; 不畏形影單,道途阻且慢,哪曲折,如渡飛湍。斬浪劈波酬壯志,同把酒,共言歡! -如夢令

            浙大校賽 B

            首先聲明 這個代碼是錯的。。。

            #include<iostream>
            #include
            <algorithm>
            #include
            <cstdio>
            using namespace std;


            int dp[100];
            int n,m;
            int dir[8][2]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,-1},{-1,1},{1,1}};

            char s[1000][1000];

            bool god(int x,int y)
            {

                
            if(x>=0&&x<n&&y>=0&&y<m)
                    
            return true;
                
            else
                    
            return false;
            }


            void input()
            {

                
            int i;
                
            int j;
                
            for(i=0;i<n;i++)
                    scanf(
            "%s",s[i]);

            }


            int v[1000][1000];
            int cnt;
            void dfs(int x,int y)
            {
                v[x][y]
            =1;
                cnt
            ++;
                
            int i;
                
            //int nx;
                
            //int ny;
                for(i=0;i<8;i++)
                
            {
                    
            int nx=x+dir[i][0];
                    
            int ny=y+dir[i][1];
                    
            if(god(nx,ny)&&s[nx][ny]=='M'&&v[nx][ny]==0)
                        dfs(nx,ny);
                }

            }



            int main()
            {
                
            int mx;
                
            int my;
                
            int i,j;
                
            while(scanf("%d%d",&n,&m)!=EOF)
                
            {
                    memset(v,
            0,sizeof(v));
                    memset(dp,
            0,sizeof(dp));
                    input();
                    
            for(i=0;i<n;i++)
                    
            {

                        
            for(j=0;j<m;j++)
                        
            {
                            cnt
            =0;
                            
            if(s[i][j]=='M'&&v[i][j]==0)
                            
            {
                                mx
            =i;
                                my
            =j;
                                dfs(i,j);
                                
            if(cnt==22)
                                    dp[
            'I'-'A']=1;
                                
            else if(cnt==24)
                                    dp[
            'L'-'A']=1;
                                
            else if(cnt==27)
                                    dp[
            'J'-'A']=1;
                                
            else if(cnt==28)
                                    dp[
            'Y'-'A']=1;
                                
            else if(cnt==31)
                                    dp[
            'Z'-'A']=1;
                                
            else if(cnt==33)
                                    dp[
            'A'-'A']=1;
                                
            else if(cnt==37)
                                    dp[
            'H'-'A']=1;
                                
            else if(cnt==44)
                                    dp[
            'E'-'A']=1;
                                
            else if(cnt==48)
                                    dp[
            'Q'-'A']=1;
                                
            else if(cnt==49)
                                    dp[
            'B'-'A']=1;
                                
            else if(cnt==34)
                                
            {
                                    
            if(s[i+1][j+1]=='M')
                                        dp[
            'T'-'A']=1;
                                    
            else
                                        dp[
            'C'-'A']=1;
                                }

                                
            else if(cnt==36)
                                
            {
                                    
            if(s[i+6][j]=='1')
                                        dp[
            'K'-'A']=1;//倒K
                                    else if(s[i+1][j+8]=='M')
                                        dp[
            'K'-'A']=1;
                                    
            else 
                                        dp[
            'F'-'A']=1


                                }

                                
            else if(cnt==38)
                                
            {
                                    
            if(s[i+2][j+2]=='M')
                                        dp[
            'G'-'A']=1;//倒G
                                    else if(s[i+1][j]=='M')
                                        dp[
            'G'-'A']=1;//正G
                                    else if(s[i+6][j]=='1')
                                        dp[
            'U'-'A']=1;
                                    
            else dp['P'-'A']=1;


                                


                                }

                                
            else if(cnt==40)
                                
            {
                                    
            if(s[i][j+5]=='M')
                                        dp[
            'O'-'A']=1;
                                    
            else if(s[i][j+3]=='M')
                                        dp[
            'V'-'A']=1;
                                    
            else if(s[i+1][j]=='M')
                                        dp[
            'W'-'A']=1;
                                    
            else 
                                        dp[
            'X'-'A']=1;
                                }

                                
            else if(cnt==42)
                                
            {

                                    
            if(s[i+1][j+3]=='M')
                                        dp[
            'N'-'A']=1;
                                    
            else
                                        dp[
            'D'-'A']=1;
                                }

                                
            else if(cnt==43)
                                
            {
                                    
            if(s[i][j+10]=='M')
                                        dp[
            'R'-'A']=1;
                                    
            else if(s[i][j+7])
                                        dp[
            'S'-'A']=1;
                                    
            else 
                                        dp[
            'R'-'A']=1;
                                }

                            }

                                



                        }

                    }


                    
            for(i=0;i<26;i++)
                    
            {
                        
            if(dp[i]==1)
                            printf(
            "%c",'A'+i);
                    }

                    printf(
            "\n");



                }

                
            return 0;
            }


            很幼稚的想數(shù)個數(shù),加判重。做到后來發(fā)現(xiàn),有點(diǎn)困難,字母相互重疊的時候有可能會判錯。應(yīng)該多思考一下再寫的。

            posted on 2010-04-10 17:56 abilitytao 閱讀(1106) 評論(0)  編輯 收藏 引用


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