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            Uriel's Corner

            Research Associate @ Harvard University / Research Interests: Computer Vision, Biomedical Image Analysis, Machine Learning
            posts - 0, comments - 50, trackbacks - 0, articles - 594

            POJ 1888 Crossword Answers---模擬

            Posted on 2009-09-03 01:48 Uriel 閱讀(315) 評(píng)論(0)  編輯 收藏 引用 所屬分類: POJ模擬
            模擬英語(yǔ)填字游戲。。關(guān)于序號(hào)處理丟了幾天突然想到。。
            最近這類模擬做了很多。。紀(jì)念下~
            /*Problem: 1888  User: Uriel 
               Memory: 292K  Time: 32MS 
               Language: C++  Result: Accepted
            */
             

            #include
            <stdio.h>
            #include
            <stdlib.h>
            #include
            <string.h>

            int g,i,j,a,b,k,x,y,syl;
            char str[20][20];

            struct M{
                
            char r[20];
                
            char c[20];
                
            int num;
                
            int flag1;
                
            int flag2;
            }
            S[2000];

            int main()
            {
                g
            =1;
                
            while(1)
                
            {
                    scanf(
            "%d",&a);
                    
            if(a==0)break;
                    scanf(
            "%d",&b);
                    
            for(i=0;i<a;i++)
                    
            {
                        getchar();
                        
            for(j=0;j<b;j++)
                        
            {
                            scanf(
            "%c",&str[i][j]);
                        }

                    }

                    x
            =1;
                    S[
            1].flag1=0;
                    S[
            1].flag2=0;
                    memset(S,
            0x00,sizeof(S));
                    
            for(i=0;i<a;i++)
                    
            {
                        
            for(j=0;j<b;j++)
                        
            {
                            syl
            =0;
                            
            if((j==0 || str[i][j-1]=='*'&& str[i][j]!='*')
                            
            {
                                syl
            =1;
                                S[x].num
            =x;
                                S[x].flag1
            =1;
                                y
            =0;
                                
            for(k=j;k<b;k++)
                                
            {
                                    
            if(str[i][k]=='*')break;
                                    S[x].r[y
            ++]=str[i][k];
                                }

                            }

                            
            if((i==0 || str[i-1][j]=='*'&& str[i][j]!='*')
                            
            {
                                syl
            =1;
                                S[x].num
            =x;
                                S[x].flag2
            =1;
                                y
            =0;
                                
            for(k=i;k<a;k++)
                                
            {
                                    
            if(str[k][j]=='*')break;
                                    S[x].c[y
            ++]=str[k][j];
                                }

                            }

                            
            if(syl)
                            
            {
                                x
            ++;
                                S[x].flag1
            =0;
                                S[x].flag2
            =0;
                            }

                        }

                    }

                    printf(
            "puzzle #%d:\n",g++);
                    printf(
            "Across\n");
                    
            for(i=0;i<x;i++)
                    
            {
                        
            if(S[i].flag1==1)
                        
            {
                            printf(
            "%3d.%s\n",S[i].num,S[i].r);
                        }

                    }

                    printf(
            "Down\n");
                    
            for(i=0;i<x;i++)
                    
            {
                        
            if(S[i].flag2==1)
                        
            {
                            printf(
            "%3d.%s\n",S[i].num,S[i].c);
                        }

                    }

                    printf(
            "\n");
                }

                system(
            "PAUSE");
                
            return 0;
            }

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