HDU 2717 Catch That Cow

该题用广搜比较好，这里要处理好就是X*2的时候，我们知道如果一步一步地走也就最多终点与起点相减的绝对值，那么我们就不能超过终点加上他们的绝对值；

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
class Node
{
public:
     int time,number;
};
Node queue[200024];
int BFS( int A, int B )
{
   bool hash[200024]={0};
   int dis = abs( A  - B );
   int     first=0,end=0;
   queue[end].time=0;
   queue[end].number = A;
   hash[A] = 1;
   end++;
   while( first < end )
   {
      Node t = queue[first];
      int x = t.number + 1;
      if( x <= B&&!hash[x] )
      {
         queue[end].number = x;
         queue[end].time = t.time + 1;
         hash[x] = 1;
         if( x == B ) return t.time + 1;
         end++;        
      }    
      x = t.number*2;
      if( x <= (B + dis)&&!hash[x] )
      {
         queue[end].number = x;
         queue[end].time = t.time + 1;
         hash[x] = 1;
         if( x == B ) return t.time + 1;
         end++;        
      }
      x = t.number -1;
      if( x >=0 && !hash[x] )
      {
         queue[end].number = x;
         queue[end].time = t.time + 1;
         hash[x] = 1;
         if( x == B ) return t.time + 1;
         end++;             
      }
      first++;    
   }
}
int main( )
{
   int A,B;
   while( scanf( "%d%d",&A,&B )==2 )
   {
       if( A == B ) printf( "0\n" );
       else 
         printf( "%d\n",BFS( A  , B ) );        
   }
   return 0;    
}