【原】 POJ 2155 Matrix 2D树状数组 解题报告

http://poj.org/problem?id=2155

题目要求修改一个区域，然后求一个元素的值。貌似和树状数组的功能完全相反。但这题，应该
说这个思路的精妙之处就体现在这里。我认为关键是要理解“树状”的概念。画一个一维的树状数
组图形，就会发现所有的元素都会链接到2,4,8,16,32……这条“主干”上来。那么，修改了“主干”，
其实就相当于修改了整个“树”。

对于每个元素a，“主干”可以分为两个部分，
Up(a) = { a1 = a, a2 = a1+ lowbit(a1), a3 = a2 + lowbit(a2), ... }
Down(a) = { a1 =a, a2 = a1 -lowbit(a1), a3 = a2 - lowbit(a2), ... }。
当然这里不是完全的对应主干上的每个元素。对于任意a<b，up(a)与down(b)的交集都只有一个元素。
若要对[a,b]内的元素进行+n操作，只要down(a-1)的元素-n, down(b)的元素+n就好了。

设a<=c<=b，查询c时则对up(c)进行求和。因为up(c)与down(b)只有一个共同元素，所以可以完整的
体现修改。以上是一维的情况。可以非常直接的推论到二维。只要对x, y坐标同时up或down。理解了
上面的思路，这题就迎刃而解了。

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N). 
We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions. 
1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2). 
2. Q x y (1 <= x, y <= n) querys A[x, y].

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case. 
The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.

Output

For each querying output one line, which has an integer representing A[x, y]. 
There is a blank line between every two continuous test cases.

Sample Input

1

2 10

C 2 1 2 2

Q 2 2

C 2 1 2 1

Q 1 1

C 1 1 2 1

C 1 2 1 2

C 1 1 2 2

Q 1 1

C 1 1 2 1

Q 2 1

Sample Output

1

0

0

1

const int N = 1001 ;

int tree[N][N] = {0} ;

//此过程是向上求的，不同于标准向下求
int GetSum( int n, int x, int y )
{
    int y1 ;
    int sum = 0 ;
    while( x<=n ) //***
    {
        y1 = y ;
        while( y1<=n )
        {
            sum += tree[x][y1] ;
            y1 += ( y1 & -y1 ) ;  //***
        }
        x += ( x & -x ) ;
    }
    return sum ;
}

//此过程是向下求的，不同于标准向上求
void AddVal( int x, int y, int val )
{
    int y1 ;
    while( x>0 )  //***
    {
        y1 = y ;
        while( y1>0 )
        {
            tree[x][y1] += val ;
            y1 -= ( y1 & -y1 ) ;  //***
        }
        x -= ( x & -x ) ;
    }
}

void run2155()
{
    int caseNum,n,t ;
    char c ;
    int x1,y1,x2,y2 ;
    int i ;

    scanf( "%d" , &caseNum ) ;
    while( caseNum-- )
    {
        scanf( "%d%d", &n,&t ) ;

        memset(tree,0,sizeof(tree)) ; //对数组赋值

        while( t-- )
        {
            scanf( "\n%c", &c ) ;
            if( c=='C' )
            {
                scanf( "%d%d%d%d", &x1,&y1,&x2,&y2 ) ;
                AddVal( x2, y2, 1 ) ;
                AddVal( x1-1, y2, -1 ) ;
                AddVal( x2, y1-1, -1 ) ;
                AddVal( x1-1, y1-1, 1 ) ;
            }
            else
            {
                scanf( "%d%d", &x1,&y1 ) ;
                printf( "%d\n", GetSum(n,x1,y1)%2 ) ;
            }
        }
        printf("\n") ;
    }
}