POJ 2155 二维线段树 书套树

    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]. 

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

ac代码：

#include <iostream>
#include <cstdio>
#include <string.h>
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;
}