poj2155--Matrix（二維樹狀數組）
Matrix
Time Limit: 3000MS
Memory Limit: 65536K
Total Submissions: 18021
Accepted: 6755
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
Source
POJ Monthly,Lou Tiancheng
二維的樹狀數組，一般的修改和查詢
對二維的樹狀數組的修改
void add(int i,int j,int d,int n)
{
int x , y ;
for(x = i ; x <= n ; x += lowbit(x))
for(y = j ; y <= n ; y += lowbit(y))
c[x][y] += d;
}
查詢
int sum(int i,int j)
{
int a = 0 , x , y ;
for(x = i ; x >= 0 ; x -= lowbit(x))
for(y = j ; y >= 0 ; y -= lowbit(y))
a += c[x][y] ;
return a ;
}
題意：給出n*n的矩陣，矩陣中的值只能為0或者1，初始值全部是0，C x1 y1 x2 y2 表示在（x1,y1）和(x2,y2)圍成的矩形中的所有值變換 ；Q x y 查詢該點當前的值，也可以認為是統計該點經過了幾次的變化；
在這個題中樹狀數組由後向前更新，每一個點的值表示由該點向前的所有點變換的次數
對於C的操作
首先 將(1,1)到（x2,y2）的所有點+1 ，代表有該點向前的矩形中的變化次數均+1；再由 (x1-1,y2)(x2,y1-1)均-1，(x1-1,y1-1)+1，平衡掉多操作的點，那麼當計算該點變換次數時，由該點向後累加，得到的總和就是該點的變換次數，如果是奇數代表1，否則是0.
#include
#include
#include
using namespace std;
int c[1010][1010] ;
int lowbit(int x)
{
return x & -x ;
}
void add(int i,int j,int d)
{
int x , y ;
for(x = i ; x > 0 ; x -= lowbit(x))
for(y = j ; y > 0 ; y -= lowbit(y))
c[x][y] += d;
}
int sum(int i,int j,int n)
{
int a = 0 , x , y ;
for(x = i ; x <= n ; x += lowbit(x))
for(y = j ; y <= n ; y += lowbit(y))
a += c[x][y] ;
return a ;
}
int main()
{
int t , tt , i , j , n , m , x1 , y1 , x2 , y2 ;
char ch ;
scanf("%d", &t);
for(tt = 1 ; tt <= t ; tt++)
{
scanf("%d %d", &n, &m);
memset(c,0,sizeof(c));
while(m--)
{
getchar();
scanf("%c", &ch);
if( ch == 'C' )
{
scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
add(x2,y2,1);
add(x1-1,y2,-1);
add(x2,y1-1,-1);
add(x1-1,y1-1,1);
}
else
{
scanf("%d %d", &x1, &y1);
printf("%d\n", sum(x1,y1,n)%2 );
}
}
if(tt != t)
printf("\n");
}
return 0;
}
