POJ 1195 Mobile phones (树状数组)

题意：给定n*n矩阵，和几种在线操作，包括对某一点(x,y)值修改，查询一个矩形(x1,y1,x2,y2)的元素和。

思路：典型的在线查询，可用树状数组实现，查询矩形和时，稍微注意以下就可以了:

　　sum(x2,y2)+sum(x1-1,y1-1)-sum(x1-1,y2)-sum(x2,y1-1);

还要注意树状数组的修改操作modify(index,delta)中的index要>0。

 

  
    

   
     
   
     #include 
   
     <
   
     iostream
   
     >
   
     
#include 
   
     <
   
     cstdio
   
     >
   
     
#include 
   
     <
   
     algorithm
   
     >
   
     
#include 
   
     <
   
     memory.h
   
     >
   
     
#include 
   
     <
   
     cmath
   
     >
   
     
#include 
   
     <
   
     bitset
   
     >
   
     
#include 
   
     <
   
     queue
   
     >
   
     
#include 
   
     <
   
     vector
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;


   
     const
   
      
   
     int
   
      BORDER 
   
     =
   
      (
   
     1
   
     <<
   
     20
   
     )
   
     -
   
     1
   
     ;

   
     const
   
      
   
     int
   
      MAXSIZE 
   
     =
   
      
   
     37
   
     ;

   
     const
   
      
   
     int
   
      MAXN 
   
     =
   
      
   
     1100
   
     ;

   
     const
   
      
   
     int
   
      INF 
   
     =
   
      
   
     1000000000
   
     ;

   
     #define
   
      CLR(x,y) memset(x,y,sizeof(x))
   
     

   
     #define
   
      ADD(x) x=((x+1)&BORDER)
   
     

   
     #define
   
      IN(x) scanf("%d",&x)
   
     

   
     #define
   
      OUT(x) printf("%d\n",x)
   
     

   
     #define
   
      MIN(m,v) (m)<(v)?(m):(v)
   
     

   
     #define
   
      MAX(m,v) (m)>(v)?(m):(v)
   
     

   
     #define
   
      ABS(x) ((x)>0?(x):-(x))
   
     


   
     int
   
      tre[MAXN][MAXN];

   
     int
   
      n,n_tre,t_case;


   
     int
   
      lowbit(
   
     int
   
      x)
{
 
   
     return
   
      x
   
     &
   
     (
   
     -
   
     x);
}

   
     void
   
      modify(
   
     const
   
      
   
     int
   
     &
   
      x,
   
     const
   
      
   
     int
   
     &
   
      y,
   
     const
   
      
   
     int
   
      delta)
{
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      x; i 
   
     <=
   
      n_tre; i 
   
     +=
   
      lowbit(i))
 
   
     for
   
     (
   
     int
   
      j 
   
     =
   
      y; j 
   
     <=
   
      n_tre; j 
   
     +=
   
      lowbit(j))
 {
 tre[i][j] 
   
     +=
   
      delta;
 }
}

   
     int
   
      get_sum(
   
     const
   
      
   
     int
   
     &
   
      x,
   
     const
   
      
   
     int
   
     &
   
      y)
{
 
   
     int
   
      sum 
   
     =
   
      
   
     0
   
     ;
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      x; i 
   
     >
   
      
   
     0
   
     ; i 
   
     -=
   
      lowbit(i))
 
   
     for
   
     (
   
     int
   
      j 
   
     =
   
      y; j 
   
     >
   
      
   
     0
   
     ; j 
   
     -=
   
      lowbit(j))
 sum 
   
     +=
   
      tre[i][j];
 
   
     return
   
      sum;
}

   
     int
   
      init()
{
 CLR(tre,
   
     0
   
     );
 
   
     return
   
      
   
     0
   
     ;
}

   
     int
   
      work()
{
 
   
     int
   
      i,j,x1,x2,y1,y2,x,y,val,tmp,ans;
 
   
     char
   
      s[
   
     3
   
     ];
 n_tre 
   
     =
   
      n
   
     +
   
     5
   
     ;
 
   
     while
   
     (scanf(
   
     "
   
     %s
   
     "
   
     ,s))
 {
 
   
     if
   
     (s[
   
     0
   
     ] 
   
     ==
   
      
   
     '
   
     3
   
     '
   
     )
 
   
     break
   
     ;
 
   
     if
   
     (s[
   
     0
   
     ] 
   
     ==
   
      
   
     '
   
     1
   
     '
   
     )
 {
 scanf(
   
     "
   
     %d%d%d
   
     "
   
     ,
   
     &
   
     x,
   
     &
   
     y,
   
     &
   
     val);
 modify(x
   
     +
   
     1
   
     ,y
   
     +
   
     1
   
     ,val);
 }
   
     else
   
      
 {
 scanf(
   
     "
   
     %d%d%d%d
   
     "
   
     ,
   
     &
   
     x1,
   
     &
   
     y1,
   
     &
   
     x2,
   
     &
   
     y2);
 ans 
   
     =
   
      get_sum(x2
   
     +
   
     1
   
     ,y2
   
     +
   
     1
   
     )
   
     +
   
     get_sum(x1,y1)
   
     -
   
     
 get_sum(x1,y2
   
     +
   
     1
   
     )
   
     -
   
     get_sum(x2
   
     +
   
     1
   
     ,y1);
 OUT(ans);
 }
 }
 
   
     return
   
      
   
     0
   
     ;
}

   
     int
   
      main()
{
 
   
     while
   
     (scanf(
   
     "
   
     %d %d
   
     "
   
     ,
   
     &
   
     t_case,
   
     &
   
     n)
   
     !=
   
     EOF)
 {
 init();
 work();
 }
 
   
     return
   
      
   
     0
   
     ;
}