HDU 2452 Navy maneuvers

HDU 2452 Navy maneuvers

JJ哥让我帮他去看这道题，我顺手A了，结果发现这题的耗时竟然排第一了。
d[i][1]表示从i点出发可以取得的最大值，d[i][0]则表示最小值，那么就有d[i][1]=max{ d[j][0] } + r[i]，d[i][0]=min{ d[j][1] } + r[i]，其中有一条<i,j>的边。
以下是我的代码：

/*
 * Author:  lee1r
 * Created Time:  2011/8/24 19:36:33
 * File Name: hdu2452.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;

bool  scanf( int   & num)
{
     char   in ;
     while (( in = getchar()) != EOF  &&  ( in > ' 9 '   ||   in < ' 0 ' ));
     if ( in == EOF)  return   false ;
    num = in - ' 0 ' ;
     while ( in = getchar(), in >= ' 0 '   &&   in <= ' 9 ' ) num *= 10 ,num += in - ' 0 ' ;
     return   true ;
}

const   int  kMaxn( 10007 );
const   int  kMaxm( 1000007 );

int  n,m,f,start,r[kMaxn],d[kMaxn][ 2 ];
int  cnt,first[kMaxn],next[kMaxm],e[kMaxm];
bool  into[kMaxn], out [kMaxn];

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

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

int  dp( int  u, int  sign)
{
     if (d[u][sign] !=- 1 )
         return  d[u][sign];
    d[u][sign] = r[u];
     int  t(sign ? 0 :kInf);
     for ( int  i = first[u];i !=- 1 ;i = next[i])
    {
         int  v(e[i]);
         if (sign)
            t = max(t,dp(v, 0 ));
         else
            t = min(t,dp(v, 1 ));
    }
    d[u][sign] += t;
     return  d[u][sign];
}

int  main()
{
    #ifndef ONLINE_JUDGE
     // freopen("data.in","r",stdin);
     #endif
    
     while (scanf(n)  &&  scanf(m)  &&  scanf(f))
    {
         for ( int  i = 1 ;i <= n;i ++ )
            scanf(r[i]);
        
        Clear();
        memset(into, false , sizeof (into));
        memset( out , false , sizeof ( out ));
         while (m -- )
        {
             int  u,v;
            scanf(u);
            scanf(v);
            AddEdge(u,v);
            into[v] = true ;
             out [u] = true ;
        }
        
        memset(d, - 1 , sizeof (d));
         for ( int  i = 1 ;i <= n;i ++ )
        {
             if ( ! out [i])
                d[i][ 0 ] = d[i][ 1 ] = r[i];
             else   if ( ! into[i])
                start = i;
        }
        
         if (dp(start, 1 ) >= f)
            printf( " Victory\n " );
         else
            printf( " Glory\n " );
    }
    
     return   0 ;
}