pku2525 Text Formalization 恶心的模拟

pku2525 Text Formalization 恶心的模拟

题意是这样：
将一篇文章中的缩写和简略语替换成原来的样子。
有点要求，缩写只在第一次替换。如果有多种替换方法，则最先出现的可替换词优先替换。如果还有重复，则列表前面的先替换。
解法：
狂用string.h里的函数然后+暴力就可以了，这种模拟题一般不会考到什么算法。。
代码：

  1  # include  < cstdio >
  2  using   namespace  std;
  3  # include  < cstring >
  4  # include  < string >
  5  char  contraction[ 201 ][ 2 ][ 100 ],acronym[ 51 ][ 2 ][ 100 ];
  6  int  main()
  7  {
  8       int  c,a;
  9      scanf( " %d%d " , & c, & a);
 10      getchar();
 11       for ( int  i = 0 ;i < c;i ++ )
 12      {
 13           char  tmp[ 200 ];
 14          gets(tmp);
 15           char   * p1 = strstr(tmp, " \ "" );
 16           char   * p2 = strstr(p1 + 1 , " \ "" );
 17           * p2 = ' \0 ' ;
 18          strcpy(contraction[ 3 * i][ 0 ],p1 + 1 );
 19          p1 = strstr(p2 + 1 , " \ "" );
 20          p2 = strstr(p1 + 1 , " \ "" );
 21           * p2 = ' \0 ' ;
 22          strcpy(contraction[ 3 * i][ 1 ],p1 + 1 );
 23          strcpy(contraction[ 3 * i + 1 ][ 0 ],contraction[ 3 * i][ 0 ]);
 24          strcpy(contraction[ 3 * i + 1 ][ 1 ],contraction[ 3 * i][ 1 ]);
 25          contraction[ 3 * i + 1 ][ 0 ][ 0 ] = (contraction[ 3 * i + 1 ][ 0 ][ 0 ] >= ' a ' && contraction[ 3 * i + 1 ][ 0 ][ 0 ] <= ' z ' ? contraction[ 3 * i + 1 ][ 0 ][ 0 ] - 32 :contraction[ 3 * i + 1 ][ 0 ][ 0 ]);
 26          contraction[ 3 * i + 1 ][ 1 ][ 0 ] = (contraction[ 3 * i + 1 ][ 1 ][ 0 ] >= ' a ' && contraction[ 3 * i + 1 ][ 1 ][ 0 ] <= ' z ' ? contraction[ 3 * i + 1 ][ 1 ][ 0 ] - 32 :contraction[ 3 * i + 1 ][ 1 ][ 0 ]);
 27          strcpy(contraction[ 3 * i + 2 ][ 0 ],contraction[ 3 * i][ 0 ]);
 28          strcpy(contraction[ 3 * i + 2 ][ 1 ],contraction[ 3 * i][ 1 ]);
 29           for ( int  j = 0 ;j < strlen(contraction[ 3 * i + 2 ][ 0 ]);j ++ )
 30              contraction[ 3 * i + 2 ][ 0 ][j] = (contraction[ 3 * i + 2 ][ 0 ][j] >= ' a ' && contraction[ 3 * i + 2 ][ 0 ][j] <= ' z ' ? contraction[ 3 * i + 2 ][ 0 ][j] - 32 :contraction[ 3 * i + 2 ][ 0 ][j]);
 31           for ( int  j = 0 ;j < strlen(contraction[ 3 * i + 2 ][ 1 ]);j ++ )
 32              contraction[ 3 * i + 2 ][ 1 ][j] = (contraction[ 3 * i + 2 ][ 1 ][j] >= ' a ' && contraction[ 3 * i + 2 ][ 1 ][j] <= ' z ' ? contraction[ 3 * i + 2 ][ 1 ][j] - 32 :contraction[ 3 * i + 2 ][ 1 ][j]);
 33      }
 34      c *= 3 ;
 35       for ( int  i = 0 ;i < a;i ++ )
 36      {
 37           char  tmp[ 200 ];
 38          gets(tmp);
 39           char   * p1 = strstr(tmp, " \ "" );
 40           char   * p2 = strstr(p1 + 1 , " \ "" );
 41           * p2 = ' \0 ' ;
 42          strcpy(acronym[i][ 0 ],p1 + 1 );
 43          p1 = strstr(p2 + 1 , " \ "" );
 44          p2 = strstr(p1 + 1 , " \ "" );
 45           * p2 = ' \0 ' ;
 46          strcpy(acronym[i][ 1 ],p1 + 1 );
 47      }
 48       char  ostr[ 50000 ];
 49       char  tstr[ 50000 ];
 50       bool  used[ 51 ];
 51      memset(used, false , sizeof (used));
 52       while (gets(ostr))
 53      {
 54           if (ostr[ 0 ] == ' # ' ) 
 55          {
 56              memset(used, false , sizeof (used));
 57              puts(ostr);
 58               continue ;
 59          }
 60           char   * str = ostr;
 61           while ( true )
 62          {
 63               char   * p = NULL, * ori, * target;
 64               int  type = 0 ;
 65               for ( int  i = 0 ;i < c;i ++ )
 66              {
 67                   char   * tmp = strstr(str,contraction[i][ 0 ]);
 68                   if (tmp != NULL && (p == NULL || tmp < p)) 
 69                  {
 70                      p = tmp;
 71                      ori = contraction[i][ 0 ];
 72                      target = contraction[i][ 1 ];
 73                      type = 0 ;
 74                  }
 75              }
 76               for ( int  i = 0 ;i < a;i ++ )
 77              {
 78                   char   * tmp = strstr(str,acronym[i][ 0 ]);
 79                   if (tmp != NULL && (p == NULL || tmp < p)) 
 80                  {
 81                      p = tmp;
 82                      ori = acronym[i][ 0 ];
 83                      target = acronym[i][ 1 ];
 84                      type = i + 1 ;
 85                  }
 86              }
 87               if (p == NULL)  break ;
 88               else
 89              {
 90                   if ( ! type)
 91                  {
 92                       * p = ' \0 ' ;
 93                      strcpy(tstr,str);
 94                      strcat(tstr,target);
 95                      strcat(tstr,p + strlen(ori));
 96                      strcpy(str,tstr);
 97                      str = p + strlen(target);
 98                  }
 99                   else
100                  {
101                      type -- ;
102                       if (used[type]) str += strlen(ori);
103                       else
104                      {
105                          used[type] = true ;
106                           * p = ' \0 ' ;
107                          strcpy(tstr,str);
108                          strcat(tstr,target);
109                          strcat(tstr, "  ( " );
110                          strcat(tstr,ori);
111                          strcat(tstr, " ) " );
112                          strcat(tstr,p + strlen(ori));
113                          strcpy(str,tstr);
114                          str = p + strlen(target) + 3 + strlen(ori);
115                      }
116                  }
117              }    
118          }
119          puts(ostr);
120      }
121       return   0 ;
122      
123  }