poj 2155 Matrix（二维线段树，树套树）
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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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题解：二维线段树。
区间修改点查询，因为是异或和，所以我们可以把当前点到跟的路径上的标记累加，就是标记不下放，但是统计时需要计算所有覆盖该点的标记。
#include
#include
#include
#include
using namespace std;
int n,m,t,ans;
struct data{
int treey[4000];
int l,r;
}treex[4000];
void buildy(int x,int now,int l,int r)
{
treex[x].treey[now]=0;
if (l==r) return;
int mid=(l+r)/2;
buildy(x,now<<1,l,mid);
buildy(x,now<<1|1,mid+1,r);
}
void build(int now,int l,int r)
{
buildy(now,1,1,n);
if (l==r) return ;
int mid=(l+r)/2;
build(now<<1,l,mid);
build(now<<1|1,mid+1,r);
}
void update(int x,int now,int l,int r,int ll,int rr)
{
if (l>=ll&&r<=rr){
treex[x].treey[now]^=1;
return;
}
int mid=(l+r)/2;
if (ll<=mid) update(x,now<<1,l,mid,ll,rr);
if (rr>mid) update(x,now<<1|1,mid+1,r,ll,rr);
}
void change(int now,int l,int r,int ll,int rr,int y1,int y2)
{
if (l>=ll&&r<=rr) {
update(now,1,1,n,y1,y2);
return;
}
int mid=(l+r)/2;
if (ll<=mid) change(now<<1,l,mid,ll,rr,y1,y2);
if (rr>mid) change(now<<1|1,mid+1,r,ll,rr,y1,y2);
}
void query1(int x,int now,int l,int r,int y)
{
ans^=treex[x].treey[now];
if (l==r) return;
int mid=(l+r)/2;
if (y<=mid) query1(x,now<<1,l,mid,y);
else query1(x,now<<1|1,mid+1,r,y);
}
void query(int now,int l,int r,int x,int y)
{
query1(now,1,1,n,y);
if (l==r) return;
int mid=(l+r)/2;
if (x<=mid) query(now<<1,l,mid,x,y);
else query(now<<1|1,mid+1,r,x,y);
}
int main()
{
freopen("a.in","r",stdin);
scanf("%d",&t);
for (int i=1;i<=t;i++)
{
scanf("%d%d",&n,&m);
build(1,1,n);
for (int j=1;j<=m;j++)
{
char s[10]; scanf("%s",s);
if (s[0]=='C'){
int x1,y1,x2,y2; scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
change(1,1,n,x1,x2,y1,y2);
}
else{
int x,y; scanf("%d%d",&x,&y);
ans=0;
query(1,1,n,x,y);
printf("%d\n",ans);
}
}
printf("\n");
}
}