POJ 1094 Sorting It All Out

POJ 1094 Sorting It All Out

题目大意：给出若干个形如A<B的表达式，判断是否出现矛盾、是否能够确定唯一的一个序列。
根据A<B构造图，出现矛盾即是出现有向环；没有矛盾就进行拓扑排序。本题需要边构图边判断，在这过程中，图可能是不连通的，通过DFS判断有向环比较方便，而用顶点的度来判断序列是否唯一比较方便。
以下是我的代码：

#include < iostream >
#include < string >
#include < cstdio >
#include < cstring >
using   namespace  std;
const   int  kMaxn( 27 );

int  n,m,g[kMaxn][kMaxn],degree[kMaxn];
int  used[kMaxn];

bool  circle( int  u)
{
    used[u] =- 1 ;
     for ( int  i = 1 ;i <= n;i ++ )
         if (g[u][i])
        {
             if (used[i] ==- 1 )
                 return   true ;
             else   if (used[i] == 0   &&  circle(i))
                 return   true ;
        }
    used[u] = 1 ;
     return   false ;
}

bool  toposort( string   & ansstr)
{
     int  d[kMaxn];

    ansstr.clear();
    memcpy(d,degree,kMaxn * sizeof ( int ));
     for ( int  i = 1 ;i <= n;i ++ )
    {
         int  now,cnt( 0 );
         for ( int  j = 1 ;j <= n;j ++ )
             if (d[j] == 0 )
            {
                now = j;
                cnt ++ ;
            }
         if (cnt > 1 )
             return   false ;
        ansstr += ( char )(now + ' A ' - 1 );
        d[now] =- 1 ;
         for ( int  j = 1 ;j <= n;j ++ )
             if (g[now][j])
                d[j] -- ;
    }
     return  (ansstr.size() == n);
}

int  main()
{
     /*
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    // */

     while (scanf( " %d%d " , & n, & m) == 2   &&  (n  ||  m))
    {
         int  anstype( - 1 ),ansvalue;
         string  ansstr;
        memset(g, 0 ,kMaxn * kMaxn * sizeof ( int ));
        memset(degree, 0 ,kMaxn * sizeof ( int ));
         for ( int  i = 1 ;i <= m;i ++ )
        {
             string  t;
            cin >> t;
            g[t[ 0 ] - ' A ' + 1 ][t[ 2 ] - ' A ' + 1 ] = 1 ;
            degree[t[ 2 ] - ' A ' + 1 ] ++ ;

            memset(used, 0 ,kMaxn * sizeof ( int ));
             if (anstype ==- 1   &&  circle(t[ 0 ] - ' A ' + 1 ))
            {
                anstype = 3 ;
                ansvalue = i;
            }
             if (anstype ==- 1   &&  toposort(ansstr))
            {
                anstype = 1 ;
                ansvalue = i;
            }
        }

         if (anstype == 3 )
            printf( " Inconsistency found after %d relations.\n " ,ansvalue);
         else   if (anstype == 1 )
            printf( " Sorted sequence determined after %d relations: %s.\n " ,ansvalue,ansstr.c_str());
         else
            printf( " Sorted sequence cannot be determined.\n " );
    }

     return   0 ;
}