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本文作者：ljh2000
作者博客：http://www.cnblogs.com/ljh2000-jump/
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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 1
正解：二维树状数组
解题报告：
推荐：2009年国家集训队论文：《浅谈信息学竞赛中的“0”和“1”》
我居然一开始没反应过来为什么只要改四个点就可以了QAQ
考虑我支持的是一个修改标记，通过一维的情况可以类比推理二维的情况，yy一下可以想清楚的。
//It is made by ljh2000
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <string>
using namespace std;
typedef long long LL;
const int MAXN = 1011;
int n,m,c[MAXN][MAXN],x[2],y[2],ans;
char ch[12];
inline void add(int x,int y){ for(int i=x;i<=n;i+=i&(-i)) for(int j=y;j<=n;j+=j&(-j)) c[i][j]++; }
inline int query(int x,int y){ int tot=0; for(int i=x;i>=1;i-=i&(-i)) for(int j=y;j>=1;j-=j&(-j)) tot+=c[i][j]; return tot; }
inline int getint(){
int w=0,q=0; char c=getchar(); while((c<'0'||c>'9') && c!='-') c=getchar();
if(c=='-') q=1,c=getchar(); while (c>='0'&&c<='9') w=w*10+c-'0',c=getchar(); return q?-w:w;
}
inline void work(){
int T=getint();
while(T--){
n=getint(); m=getint();
memset(c,0,sizeof(c));
while(m--) {
scanf("%s",ch);
if(ch[0]=='C') {
x[0]=getint(); y[0]=getint();
x[1]=getint(); y[1]=getint();
add(x[0],y[0]);
add(x[0],y[1]+1);
add(x[1]+1,y[0]);
add(x[1]+1,y[1]+1);
}
else{
x[0]=getint(); y[0]=getint();
ans=query(x[0],y[0]);
printf("%d
",ans&1);
}
}
printf("
");
}
}
int main()
{
work();
return 0;
}