ZOJ 3129 || POJ 3067 Japan

同POJ 2299 也是利用树状数组求逆序数的应用

将x从大到小排序，若x相同，按y从大到小排序，对y建立树状数组，根据逆序数的定义，画图演算一遍就很清楚了

  
    

   
     #include 
   
     <
   
     stdio.h
   
     >
   
     
#include 
   
     <
   
     stdlib.h
   
     >
   
     
#include 
   
     <
   
     string
   
     .h
   
     >
   
     

   
     #define
   
      MAXN 1001
   
     

   
     struct
   
      Pair{
 
   
     int
   
      x, y;
}p[
   
     1000010
   
     ];

   
     int
   
      n, m, k;

   
     int
   
      c[MAXN];


   
     int
   
      cmp(
   
     const
   
      
   
     void
   
      
   
     *
   
     a, 
   
     const
   
      
   
     void
   
      
   
     *
   
     b){
 
   
     if
   
     (((Pair
   
     *
   
     )a)
   
     ->
   
     x 
   
     ==
   
      ((Pair
   
     *
   
     )b)
   
     ->
   
     x)
 
   
     return
   
      ((Pair
   
     *
   
     )b)
   
     ->
   
     y 
   
     -
   
      ((Pair
   
     *
   
     )a)
   
     ->
   
     y; 
 
   
     return
   
      ((Pair
   
     *
   
     )b)
   
     ->
   
     x 
   
     -
   
      ((Pair
   
     *
   
     )a)
   
     ->
   
     x;
}
inline 
   
     int
   
      lowbit(
   
     int
   
      t){
 
   
     return
   
      t 
   
     &
   
      (
   
     -
   
     t);
}

   
     void
   
      update(
   
     int
   
      i){
 
   
     while
   
     (i 
   
     <=
   
      m){
 c[i] 
   
     +=
   
      
   
     1
   
     ; 
   
     //
   
     增量为1
   
     

   
      i 
   
     +=
   
      lowbit(i);
 }
}

   
     int
   
      getsum(
   
     int
   
      i){
 
   
     int
   
      sum 
   
     =
   
      
   
     0
   
     ;
 
   
     while
   
     (i 
   
     >
   
      
   
     0
   
     ){
 sum 
   
     +=
   
      c[i];
 i 
   
     -=
   
      lowbit(i);
 }
 
   
     return
   
      sum;
}

   
     int
   
      main(){
 
   
     int
   
      T, t 
   
     =
   
      
   
     0
   
     ;
 scanf(
   
     "
   
     %d
   
     "
   
     , 
   
     &
   
     T);
 
   
     while
   
     ( T
   
     --
   
      ){
 scanf(
   
     "
   
     %d%d%d
   
     "
   
     ,
   
     &
   
     n, 
   
     &
   
     m, 
   
     &
   
     k);
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <=
   
      k; 
   
     ++
   
     i){
 scanf(
   
     "
   
     %d%d
   
     "
   
     ,
   
     &
   
     p[i].x, 
   
     &
   
     p[i].y);
 }
 qsort(p 
   
     +
   
      
   
     1
   
     , k, 
   
     sizeof
   
     (p[
   
     0
   
     ]), cmp);
 
   
     long
   
      
   
     long
   
      ans 
   
     =
   
      
   
     0
   
     ;
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <=
   
      k; 
   
     ++
   
     i){
 ans 
   
     +=
   
      getsum(p[i].y 
   
     -
   
      
   
     1
   
     );
 update(p[i].y);
 }
 printf(
   
     "
   
     Test case %d: %lld\n
   
     "
   
     ,
   
     ++
   
     t, ans);
 memset(c, 
   
     0
   
     , 
   
     sizeof
   
     (c));
 }
 
   
     return
   
      
   
     0
   
     ;
}