POJ 3715 Blue and Red

POJ 3715 Blue and Red

题目就是让求一个二分图的最小点覆盖，然后输出字典序最小的方案。
依次从二分图中删除0..N-1中的一个点，如果此时最大匹配数减少了，就永久删除这个点，然后输出这个点，否则将该点恢复。
使用Hopcroft Karp算法157ms。
以下是我的代码：

/*
 * Author:  lee1r
 * Created Time:  2011/8/19 21:40:58
 * File Name: poj3715.cpp
  */
#include < iostream >
#include < sstream >
#include < fstream >
#include < vector >
#include < list >
#include < deque >
#include < queue >
#include < stack >
#include < map >
#include < set >
#include < bitset >
#include < algorithm >
#include < cstdio >
#include < cstdlib >
#include < cstring >
#include < cctype >
#include < cmath >
#include < ctime >
#define  L(x) ((x)<<1)
#define  R(x) ((x)<<1|1)
#define  Half(x) ((x)>>1)
#define  Lowbit(x) ((x)&(-(x)))
using   namespace  std;
const   int  kInf( 0x7f7f7f7f );
const   double  kEps(1e - 8 );
typedef unsigned  int   uint ;
typedef  long   long  int64;
typedef unsigned  long   long  uint64;

const   int  kMaxn( 207 );
const   int  kMaxm( 20007 );

struct  Edge
{
     int  u,v;
};

int  N,M,r[kMaxn];
int  cnt,first[kMaxn],next[kMaxm];Edge e[kMaxm];
int  maxmatch,cx[kMaxn],cy[kMaxn];
int  distx[kMaxn],disty[kMaxn];
int  head,tail,q[kMaxn];
bool  used[kMaxn];

void  Clear()
{
    cnt = 0 ;
    memset(first, - 1 , sizeof (first));
}

void  AddEdge( int  u, int  v)
{
    cnt ++ ;
    e[cnt].u = u;
    e[cnt].v = v;
    next[cnt] = first[u];
    first[u] = cnt;
}

bool  BFS()
{
     bool  re( false );
    head = tail = 0 ;
    memset(distx, 0 , sizeof (distx));
    memset(disty, 0 , sizeof (disty));
     for ( int  i = 0 ;i < N;i ++ )
         if (used[i]  &&   ! r[i]  &&  cx[i] ==- 1 )
            q[tail ++ ] = i;
     while (head != tail)
    {
         int  h,t;
         for (h = head,t = tail;h != t;h = (h + 1 ) % kMaxn)
        {
             int  u(q[h]);
             for ( int  i = first[u];i !=- 1 ;i = next[i])
            {
                 int  v(e[i].v);
                 if ( ! used[v])
                     continue ;
                 if ( ! disty[v])
                {
                    disty[v] = distx[u] + 1 ;
                     if (cy[v] ==- 1 )
                        re = true ;
                     else
                    {
                        distx[cy[v]] = disty[v] + 1 ;
                        q[tail] = cy[v];
                        tail = (tail + 1 ) % kMaxn;
                    }
                }
            }
        }
        head = t;
    }
     return  re;
}

bool  DFS( int  u)
{
     for ( int  i = first[u];i !=- 1 ;i = next[i])
    {
         int  v(e[i].v);
         if ( ! used[v])
             continue ;
         if (disty[v] == distx[u] + 1 )
        {
            disty[v] = 0 ;
             if (cy[v] ==- 1   ||  DFS(cy[v]))
            {
                cx[u] = v;
                cy[v] = u;
                 return   true ;
            }
        }
    }
     return   false ;
}

void  HopcroftKarp()
{
    maxmatch = 0 ;
    memset(cx, - 1 , sizeof (cx));
    memset(cy, - 1 , sizeof (cy));
    
     while (BFS())
         for ( int  i = 0 ;i < N;i ++ )
             if (used[i]  &&   ! r[i]  &&  cx[i] ==- 1   &&  DFS(i))
                maxmatch ++ ;
}

int  main()
{
    #ifndef ONLINE_JUDGE
     // freopen("data.in","r",stdin);
     #endif
    
     int  T;
    scanf( " %d " , & T);
     while (T -- )
    {
        Clear();
        
        scanf( " %d%d " , & N, & M);
         for ( int  i = 0 ;i < N;i ++ )
            scanf( " %d " , & r[i]);
         while (M -- )
        {
             int  u,v;
            scanf( " %d%d " , & u, & v);
             if (r[u] != r[v])
            {
                 if (r[u] == 0 )
                    AddEdge(u,v);
                 else
                    AddEdge(v,u);
            }
        }
        
         for ( int  i = 0 ;i < N;i ++ )
            used[i] = true ;
        HopcroftKarp();
         int  ans(maxmatch);
        printf( " %d " ,ans);
         for ( int  i = 0 ;i < N  &&  ans;i ++ )
        {
            used[i] = false ;
            HopcroftKarp();
             if (maxmatch < ans)
                printf( "  %d " ,i);
             else
                used[i] = true ;
            ans = maxmatch;
        }
        printf( " \n " );
    }
    
     return   0 ;
}