poj 3302 Subsequence

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5303		Accepted: 3162

  这一题相当得简单只要正反比较两次即可，运用队列表示正向的，用栈表示反向的

代码：

 

  
    

   
     
   
      1 
   
     #include
   
     <
   
     stdio.h
   
     >
   
     

   
      2 
   
     #include
   
     <
   
     queue
   
     >
   
     

   
      3 
   
     #include
   
     <
   
     stack
   
     >
   
     

   
      4 
   
      
   
     using
   
      
   
     namespace
   
      std;

   
      5 
   
      
   
     int
   
      main()

   
      6 
   
     {

   
      7 
   
      
   
     int
   
      t,i,mark;

   
      8 
   
      
   
     char
   
      s1[
   
     105
   
     ],s2[
   
     105
   
     ];

   
      9 
   
      

   
     10 
   
      scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     t);

   
     11 
   
      
   
     while
   
     (t
   
     --
   
     )

   
     12 
   
      {

   
     13 
   
      mark
   
     =
   
     0
   
     ;

   
     14 
   
      scanf(
   
     "
   
     %s%s
   
     "
   
     ,s1,s2);

   
     15 
   
      queue
   
     <
   
     char
   
     >
   
      qu;

   
     16 
   
      stack
   
     <
   
     char
   
     >
   
      st;

   
     17 
   
      
   
     for
   
     (i
   
     =
   
     0
   
     ;s2[i]
   
     !=
   
     '
   
     \0
   
     '
   
     ;i
   
     ++
   
     )

   
     18 
   
      {

   
     19 
   
      st.push(s2[i]);

   
     20 
   
      qu.push(s2[i]);

   
     21 
   
      }

   
     22 
   
      
   
     for
   
     (i
   
     =
   
     0
   
     ;s1[i]
   
     !=
   
     '
   
     \0
   
     '
   
     ;i
   
     ++
   
     )

   
     23 
   
      {

   
     24 
   
      
   
     char
   
      x,y;

   
     25 
   
      x
   
     =
   
     st.top();

   
     26 
   
      y
   
     =
   
     qu.front();

   
     27 
   
      
   
     if
   
     (x
   
     ==
   
     s1[i])

   
     28 
   
      st.pop();

   
     29 
   
      
   
     if
   
     (st.empty())

   
     30 
   
      {

   
     31 
   
      mark
   
     =
   
     1
   
     ;

   
     32 
   
      
   
     break
   
     ;

   
     33 
   
      }

   
     34 
   
      
   
     if
   
     (y
   
     ==
   
     s1[i])

   
     35 
   
      qu.pop();

   
     36 
   
      
   
     if
   
     (qu.empty())

   
     37 
   
      {

   
     38 
   
      mark
   
     =
   
     1
   
     ;

   
     39 
   
      
   
     break
   
     ;

   
     40 
   
      }

   
     41 
   
      }

   
     42 
   
      
   
     if
   
     (mark)

   
     43 
   
      printf(
   
     "
   
     YES\n
   
     "
   
     );

   
     44 
   
      
   
     else
   
     

   
     45 
   
      printf(
   
     "
   
     NO\n
   
     "
   
     );

   
     46 
   
      }

   
     47 
   
      
   
     return
   
      
   
     0
   
     ;

   
     48 
   
     }

   
     49