http://poj.org/problem?id=2155
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Matrix
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
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谨以此题来祭我的第一道二维线段树！
没有30道线段树的基础不要碰二维线段树。
题意：
给出一个N×N的零矩阵。有两种操作：C x1 y1 x2 y2 是将(x1,y1) (x2,y2)相应范围内的矩阵中的0/1互换,Q x y是查询(x,y)位置元素的值。
分析：
0/1互换即异或操作，朴素的算法时间复杂度显然不够。假设是一维的坐标轴显然能够用线段树解，但这是一个二维的。那么就要用到二维线段树了。
以前的我天真地觉得二维线段树就是把一个矩形直接分成4份：左上、右上、左下、右下，后来遇到一个三维的题目我还YY出了把立方体分成8份来解（当然没过啦，并且标程也不是用线段树的）。
实际上，二维线段树是先将X坐标二分，然后再将Y轴二分。
第二维维护的也是一棵线段树，没错。二维线段树就是线段树套线段树！二分。再二分！
本题每次更新的时候先操作X轴，操作到相应区间时再对Y轴进行操作，从二维到一维。达到降维的效果。
查询时每次查到在所在的X区间内时都要对Y轴进行查询。
假设对线段树理解较深。相信对此不难理解。
详情请见代码（个人习惯维护左闭右开的区间，下标从0開始，所以有些调整）：
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<cctype>
#include<cmath>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define maxm
#define maxn 1007
using namespace std;
int XOR[maxn<<2][maxn<<2];
int n,m;
//update_1d() 更新一维
//[a,b)表示当前在一维操作的区间(Y方向)。k为节点标号，[l,r)表示节点相应维护的区间(Y方向)，k_2d表示二维相应的节点标号
void update_1d(int a,int b,int k,int l,int r,int k_2d)
{
if (b<=l || r<=a) return ;
if (a<=l && r<=b)
{
XOR[k_2d][k]^=1;
}
else
{
update_1d(a,b,k*2+1,l,l+r>>1,k_2d);
update_1d(a,b,k*2+2,l+r>>1,r,k_2d);
}
}
//update_2d() 更新二维
//[a,b)表示当前在二维操作的区间(X方向)。[y1,y2)表示将要在一维操作的区间(Y方向)，k为节点标号，[l,r)表示节点相应维护的区间(X方向)
void update_2d(int a,int b,int y1,int y2,int k,int l,int r)
{
if (b<=l || r<=a) return ;
if (a<=l && r<=b)
{
update_1d(y1,y2,0,0,n,k);
}
else
{
update_2d(a,b,y1,y2,k*2+1,l,l+r>>1);
update_2d(a,b,y1,y2,k*2+2,l+r>>1,r);
}
}
//query_1d() 查询一维
//p表示当前在一维查询的位置(Y方向)，k为节点标号，[l,r)表示节点相应维护的区间(Y方向)，k_2d表示二维相应的节点标号
int query_1d(int p,int k,int l,int r,int k_2d)
{
if (r-l==1) return XOR[k_2d][k];
int m=l+r>>1;
if (p<m)
return XOR[k_2d][k]^query_1d(p,k*2+1,l,m,k_2d);
else
return XOR[k_2d][k]^query_1d(p,k*2+2,m,r,k_2d);
}
//query_2d() 查询二维
//px表示当前在二维查询的位置(X方向)，py表示将要在一维查询的位置(Y方向)，k为节点标号。[l,r)表示节点相应维护的区间(X方向)
int query_2d(int px,int py,int k,int l,int r)
{
int res=query_1d(py,0,0,n,k);// attention please!
if (r-l==1) return res;
int m=l+r>>1;
if (px<m)
return res^query_2d(px,py,k*2+1,l,m);
else
return res^query_2d(px,py,k*2+2,m,r);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("/home/fcbruce/文档/code/t","r",stdin);
#endif // ONLINE_JUDGE
int T_T;
scanf("%d",&T_T);
while (T_T--)
{
scanf("%d %d",&n,&m);
memset(XOR,0,sizeof XOR);
for (int i=0;i<m;i++)
{
char op;
int x1,x2,y1,y2;
getchar();
op=getchar();
if (op=='C')
{
scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
x1--;y1--;
update_2d(x1,x2,y1,y2,0,0,n);
}
else
{
scanf("%d %d",&x1,&y1);
x1--;y1--;
printf("%d
",query_2d(x1,y1,0,0,n));
}
}
if (T_T) putchar('
'); //由于这个PE一发
}
return 0;
}
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