Ural 1002. Phone numbers

Ural 1002. Phone numbers

/*
0.062     1485 KB
*/
#include < iostream >
using   namespace  std;

int  N;
const   char   * d = " 22233344115566070778889990 " ;
char  a[ 110 ];
char  s[ 50001 ][ 55 ];
int  ans[ 110 ];
int  f[ 110 ];
bool  check( int  x, int  y){
     int  len = strlen(s[y]);
     for ( int  i = 0 ;i < len; ++ i)
         if (a[x - i] != d[s[y][len - i - 1 ] - ' a ' ])
             return   0 ;
     return   1 ;
}
void   out ( int  len){
     if ( ! len)  return ;
     out (len - strlen(s[f[len]]));
     if (len == strlen(a)) 
        printf( " %s\n " ,s[f[len]]);
     else
        printf( " %s  " ,s[f[len]]);
}
int  main()
{
     while ( 1 ){
        gets(a);
         if (strcmp(a, " -1 " ) == 0 )  break ;
        scanf( " %d " , & N); getchar();
         for ( int  i = 0 ;i < N; ++ i)
            gets(s[i]);
        memset(ans, 0x7F , sizeof (ans));
        memset(f, - 1 , sizeof (f));
        ans[ 0 ] = 0 ;
        f[ 0 ] = 0 ;
         for ( int  i = 0 ;i < strlen(a); ++ i)
             for ( int  j = 0 ;j < N; ++ j)
                 if (i + 1 >= strlen(s[j]) && a[i] == d[s[j][strlen(s[j]) - 1 ] - ' a ' ])
                     if (ans[i + 1 ] > ans[i + 1 - strlen(s[j])] && check(i,j)){
                        ans[i + 1 ] = ans[i + 1 - strlen(s[j])] + 1 ;
                        f[i + 1 ] = j;
                    }
         if (f[strlen(a)] ==- 1 ) printf( " No solution.\n " );
         else   out (strlen(a));
    }
     return   0 ;
}

解法二：

/*
0.046
*/
#include < iostream >
using   namespace  std;

int  N;
const   char   * d = " 22233344115566070778889990 " ;
char  map[ 110 ][ 110 ][ 256 ];
char  sol[ 110 ][ 256 ];
char  s[ 256 ];
int  f[ 110 ];
void  check( int  x, char   * y){
     for ( int  i = 0 ;i < strlen(y); ++ i)
         if (s[x - i - 1 ] != d[y[strlen(y) - i - 1 ] - ' a ' ])
             return ;
    strcpy(map[x][strlen(y)],y);
}
int  main()
{
    gets(s);
    scanf( " %d " , & N); getchar();
     for ( int  i = 0 ;i <= strlen(s); ++ i){
        sol[i][ 0 ] = 0 ;
         for ( int  j = 0 ;j <= strlen(s); ++ j)
            map[i][j][ 0 ] = 0 ;
    }
     char  ch[ 255 ];
     for ( int  i = 1 ;i <= N; ++ i){
        gets(ch);
         for ( int  j = strlen(ch);j <= strlen(s); ++ j)
            check(j,ch);
    }
    f[ 0 ] = 0 ;
     for ( int  i = 1 ;i <= strlen(s); ++ i){
         if (strlen(map[i][i]) == 0 ){
            f[i] = 256 ;
            strcpy(ch,sol[i]);
             for ( int  j = 1 ;j < i; ++ j)
                 if (f[i] > f[j] + 1 && strlen(map[i][i - j]) > 0 ){
                    f[i] = f[j] + 1 ;
                    strcpy(ch,sol[i]);
                    strcat(ch,sol[j]);
                    strcat(ch, "   " );
                    strcat(ch,map[i][i - j]);
                }
            strcpy(sol[i],ch);
        }
         else {
            f[i] = 1 ;
            strcpy(sol[i],map[i][i]);
        }
    }
     if (strlen(sol[strlen(s)]) > 0 )
        printf( " %s\n " ,sol[strlen(s)]);
     else
        printf( " No solution.\n " );
     return   0 ;
}