福建农大 1084 检查站

单调队列 见 http://www.cnblogs.com/lxf90/articles/2012016.html

第一，原序列的坐标单调

第二，原序列的值单调

  
    

   
     /*
   
     单调队列 
   
     */
   
     
#include 
   
     <
   
     stdio.h
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;

   
     const
   
      
   
     int
   
      MAXN 
   
     =
   
      
   
     100001
   
     ;

   
     struct
   
      dqueue{
 
   
     int
   
      val;
 
   
     int
   
      idx;
}que[MAXN];

   
     int
   
      a[MAXN], cnt;

   
     int
   
      head, tail;

   
     int
   
      lefts;

   
     int
   
      main(){
 
   
     int
   
      n;
 
   
     //
   
     cin >> n;
   
     

   
      scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     n);
 head 
   
     =
   
      tail 
   
     =
   
      
   
     0
   
     ;
 cnt 
   
     =
   
      
   
     0
   
     ;
 lefts 
   
     =
   
      
   
     0
   
     ;
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      n; 
   
     ++
   
     i){
 
   
     //
   
     string s;
 
   
     //
   
     cin >> s;
   
     

   
      
   
     char
   
      s[
   
     20
   
     ];
 scanf(
   
     "
   
     %s
   
     "
   
     ,s);
 
   
     if
   
     (s[
   
     0
   
     ] 
   
     ==
   
      
   
     '
   
     Q
   
     '
   
     ){
 
   
     if
   
     ( head 
   
     >
   
      tail ){
 
   
     //
   
     cout << "0" << endl; 
   
     

   
      printf(
   
     "
   
     0\n
   
     "
   
     );
 } 
   
     else
   
      {
 
   
     //
   
     cout <<que[head].val << endl; 
   
     

   
      printf(
   
     "
   
     %d\n
   
     "
   
     ,que[head].val);
 }
 } 
   
     else
   
      
   
     if
   
     (s[
   
     0
   
     ] 
   
     ==
   
      
   
     '
   
     I
   
     '
   
     ){
 
   
     //
   
     cin >> a[cnt++];
   
     

   
      scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     a[cnt
   
     ++
   
     ]);
 
   
     if
   
     (head 
   
     >
   
      tail){
 head 
   
     =
   
      tail 
   
     =
   
      
   
     0
   
     ;
 que[head].idx 
   
     =
   
      cnt 
   
     -
   
      
   
     1
   
     ;
 que[head].val 
   
     =
   
      a[cnt 
   
     -
   
      
   
     1
   
     ];
 
   
     continue
   
     ; 
 }
 
   
     while
   
      (head 
   
     <=
   
      tail 
   
     &&
   
      que[tail].val 
   
     <=
   
      a[cnt 
   
     -
   
      
   
     1
   
     ]) 
 tail
   
     --
   
     ;
 
   
     ++
   
     tail;
 que[tail].val 
   
     =
   
      a[cnt 
   
     -
   
      
   
     1
   
     ];
 que[tail].idx 
   
     =
   
      cnt 
   
     -
   
      
   
     1
   
     ;
 } 
   
     else
   
      {
 lefts
   
     ++
   
     ;
 
   
     if
   
     (que[head].idx 
   
     <
   
      lefts)
 head
   
     ++
   
     ; 
 }
 }
 
   
     return
   
      
   
     0
   
     ;
}