POJ 3080 Blue Jeans

//本题爆搜即可......
//含有一点是要按字典序排序，输出最小值
//程序写得太麻烦了.....

 

code

#include < iostream >
using   namespace  std;

int  sub( char *  a, char *  b, int  n)
{
     int  i,j;
     for ( i = 0 ;i <= 60 - n; ++ i)
    {
         for ( j = 0 ;j < n; ++ j)
             if (a[j] != b[j + i])
                 break ;
         if (j == n)
             return   1 ;
    }

     return   0 ;
} // 是否字串
int  main()
{
     char  a[ 61 ],b[ 15 ][ 61 ],temp[ 61 ],tag[ 100 ];
     int  i,j,total,n,index,k,t,num,tag1,tag2,tag3;
    cin >> total;
     while (total -- )
    {
        cin >> n;
        cin >> a;
         for (i = 0 ;i < n - 1 ; ++ i)
            cin >> b[i];
         for (i = 60 ;i > 2 ; -- i)
        {
            index = 0 ;
            num = 0 ;
            tag3 = 0 ;
             for (j = 0 ;j <= 60 - i; ++ j)
            {
                tag1 = 0 ;
                tag2 = 0 ;
                 for (k = 0 ;k < n - 1 ; ++ k)
                {
                     if (sub( & a[j],b[k],i))
                        ;
                     else   break ;
                }
                 if (k == n - 1 )
                {
                    tag3 = 1 ; // 有公共
                     if (tag1 == 1 )
                    {
                        tag2 = 0 ;
                         for (t = 0 ;t < i; ++ t)
                             if (a[j + t] < temp[t])
                            {
                                tag2 = 1 ;
                                 break ;
                            }
                             else   if (a[j + t] > temp[t])
                                 break ;
                         if (tag2 == 1 )
                             for (t = 0 ;t < i; ++ t)
                                temp[t] = a[j + t];
                    }
                     else
                    {
                        tag1 = 1 ;
                         for (t = 0 ;t < i; ++ t)
                            temp[t] = a[j + t];
                    }
                }

            }
            
             if (tag3)
            {
                num = i;
                i =- 1 ;
                 break ;
            }
        }
         if (num == 0 )
            cout << " no significant commonalities " ;
         for (t = 0 ;t < num; ++ t)
            cout << temp[t];
        cout << endl;

        
    }
                

}