是一道裸的二维线段树题目，二维线段树可以用树套树的方式实现。。。。题目：
Matrix
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 11798 | Accepted: 4466 |
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].
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.
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.
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 1ac代码：
#include
#include
#include
using namespace std;
const int N=1010;
int n,m,ans;
struct newtree{
int left,right,value;
int getnewmid(){
return (left+right)/2;
}
};
struct tree{
int left,right;
newtree newtt[4*N];
int getmid(){
return (left+right)/2;
}
}tt[4*N];
void built_treey(int lp,int rp,int posx,int posy){
tt[posx].newtt[posy].left=lp;
tt[posx].newtt[posy].right=rp;
tt[posx].newtt[posy].value=0;
if(lp==rp)return;
int mid=tt[posx].newtt[posy].getnewmid();
built_treey(lp,mid,posx,posy*2);
built_treey(mid+1,rp,posx,posy*2+1);
}
void built_treex(int lp,int rp,int pos){
tt[pos].left=lp;
tt[pos].right=rp;
built_treey(1,n,pos,1);
if(lp==rp)return;
int mid=tt[pos].getmid();
built_treex(lp,mid,pos*2);
built_treex(mid+1,rp,pos*2+1);
}
void update_y(int y1,int y2,int posx,int posy){
if(tt[posx].newtt[posy].left==y1&&tt[posx].newtt[posy].right==y2){
tt[posx].newtt[posy].value=!tt[posx].newtt[posy].value;
return;
}
int mid=tt[posx].newtt[posy].getnewmid();
if(y2<=mid)
update_y(y1,y2,posx,posy*2);
else if(y1>mid)
update_y(y1,y2,posx,posy*2+1);
else{
update_y(y1,mid,posx,posy*2);
update_y(mid+1,y2,posx,posy*2+1);
}
}
void update_x(int x1,int x2,int y1,int y2,int pos){
if(tt[pos].left==x1&&tt[pos].right==x2){
update_y(y1,y2,pos,1);
return;
}
int mid=tt[pos].getmid();
if(x2<=mid)
update_x(x1,x2,y1,y2,pos*2);
else if(x1>mid)
update_x(x1,x2,y1,y2,pos*2+1);
else{
update_x(x1,mid,y1,y2,pos*2);
update_x(mid+1,x2,y1,y2,pos*2+1);
}
}
void find_y(int y,int posx,int posy){
ans^=tt[posx].newtt[posy].value;
if(tt[posx].newtt[posy].left==tt[posx].newtt[posy].right)
return;
int mid=tt[posx].newtt[posy].getnewmid();
if(y>mid)
find_y(y,posx,posy*2+1);
else
find_y(y,posx,posy*2);
}
void find_x(int x,int y,int pos){
find_y(y,pos,1);
if(tt[pos].left==tt[pos].right)
return;
int mid=tt[pos].getmid();
if(x>mid)
find_x(x,y,pos*2+1);
else
find_x(x,y,pos*2);
}
int main(){
//freopen("1.txt","r",stdin);
int numcase;
scanf("%d",&numcase);
while(numcase--){
scanf("%d%d",&n,&m);
built_treex(1,n,1);
char ss[2];
int x1,x2,y1,y2,x,y;
while(m--){
scanf("%s",ss);
if(ss[0]=='C'){
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
update_x(x1,x2,y1,y2,1);
}
else{
scanf("%d%d",&x,&y);
ans=0;
find_x(x,y,1);
printf("%d\n",ans);
}
}
puts("");
}
return 0;
}