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  
# include<iostream>
# include<cstdio>
# include<cstring>
# include<string>
# include<set>
# include<map>
# include<sstream>
using namespace std;
const int maxn=1000+5;
int tree[maxn][maxn];
char a[10];
int n;
int lowbit(int x)
{
return x&-x;
}
int getsum(int x,int y)
{
int s=0;
for(int i=x;i>0;i-=lowbit(i))
for(int j=y;j>0;j-=lowbit(j))
s+=tree[i][j];
if(s%2==0) return 0;
else
return 1;
}
void update(int x,int y)
{
for(int i=x;i<=n;i+=lowbit(i))
for(int j=y;j<=n;j+=lowbit(j))
tree[i][j]^=1;
}
int main()
{
/*#ifndef ONLINE_JUDGE
freopen("in.cpp","r",stdin);
freopen("out.cpp","w",stdout);
#endif*/
int t;
scanf("%d",&t);
while(t--)
{
int m;
scanf("%d%d",&n,&m);
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
tree[i][j]=0;
while(m--)
{
int x1,y1,x2,y2;
scanf("%s%d%d",&a[0],&x1,&y1);
if(a[0]=='C')
{
scanf("%d%d",&x2,&y2);
update(x2+1,y2+1);
update(x2+1,y1);
update(x1,y2+1);
update(x1,y1);
}
else if(a[0]=='Q')
{
printf("%d\n",getsum(x1,y1));
}
}
cout<<endl;
} return 0;
}

Matrix+POJ+二维树状数组初步的更多相关文章

  1. POJ 2155 Matrix【二维树状数组+YY(区间计数)】

    题目链接:http://poj.org/problem?id=2155 Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissio ...

  2. POJ 2155 Matrix (二维树状数组)

    Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 17224   Accepted: 6460 Descripti ...

  3. POJ 2155 Matrix(二维树状数组+区间更新单点求和)

    题意:给你一个n*n的全0矩阵,每次有两个操作: C x1 y1 x2 y2:将(x1,y1)到(x2,y2)的矩阵全部值求反 Q x y:求出(x,y)位置的值 树状数组标准是求单点更新区间求和,但 ...

  4. POJ 2155 Matrix 【二维树状数组】(二维单点查询经典题)

    <题目链接> 题目大意: 给出一个初始值全为0的矩阵,对其进行两个操作. 1.给出一个子矩阵的左上角和右上角坐标,这两个坐标所代表的矩阵内0变成1,1变成0. 2.查询某个坐标的点的值. ...

  5. POJ 2155 Matrix (二维树状数组)题解

    思路: 没想到二维树状数组和一维的比只差了一行,update单点更新,query求和 这里的函数用法和平时不一样,query直接算出来就是某点的值,怎么做到的呢? 我们在更新的时候不止更新一个点,而是 ...

  6. [POJ2155]Matrix(二维树状数组)

    题目:http://poj.org/problem?id=2155 中文题意: 给你一个初始全部为0的n*n矩阵,有如下操作 1.C x1 y1 x2 y2 把矩形(x1,y1,x2,y2)上的数全部 ...

  7. POJ2155 Matrix(二维树状数组||区间修改单点查询)

    Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row an ...

  8. HLJU 1188 Matrix (二维树状数组)

    Matrix Time Limit: 4 Sec  Memory Limit: 128 MB Description 给定一个1000*1000的二维矩阵,初始矩阵中每一个数都为1,然后为矩阵有4种操 ...

  9. PKU 2155 Matrix(裸二维树状数组)

    题目大意:原题链接 题意很简单,就不赘诉了. 解题思路: 使用二维树状数组,很裸的题. 二维的写起来也很方便,两重循环. Add(int x,int y,int val)表示(x,y)-(n,n)矩形 ...

随机推荐

  1. Linux pipe功能

    1. 功能说明 pipe(管道建设): 1) 头 #include<unistd.h> 2) 定义函数: int pipe(int filedes[2]); 3) 函数说明: pipe() ...

  2. 解压tar.gz文件报错gzip: stdin: not in gzip format解决方法

    解压tar.gz文件报错gzip: stdin: not in gzip format解决方法 在解压tar.gz文件的时候报错 1 2 3 4 5 [Sun@localhost Downloads] ...

  3. Linux下的softlink和hardlink(转)

    Linux中包括两种链接:硬链接(hard link)和软链接(soft link),软链接又称为符号链接(symbolic link) 创建命令:ln -s destfile/directory s ...

  4. Java Web----Java Web的数据库操作(二)

    Java Web的数据库操作 三.JDBC操作数据库 上一篇介绍了JDBC API,之后就可以通过API来操作数据库,实现对数据库的CRUD操作了. http://blog.csdn.net/zhai ...

  5. [linux]ubuntu14.04通过apt-get安装软件失败

    1.首先查看 dns 配置 sudo vi /etc/resolv.conf nameserver 114.114.114.114 nameserver 8.8.8.8 2.修改 apt-get 源 ...

  6. wpf dll和exe合并成一个新的exe

    原文:wpf dll和exe合并成一个新的exe 微软有一个工具叫ILMerge可以合并dll exe等,但是对于wpf的应用程序而言这个工具就不好用了.我的这方法也是从国外一个博客上找来的.仅供大家 ...

  7. 自顶向下分析Binder【1】—— Binder实例篇

    欢迎转载,转载请注明:http://blog.csdn.net/zhgxhuaa 一个Binder实例 我们Binder的学习将从以下的一个实例開始.依据Android文档中的描写叙述,创建一个Bin ...

  8. CC ANUMLA(STL的运用)

    题目连接:http://www.codechef.com/problems/ANUMLA 题意:给一个序列所有子集和(2^n个子集),复原这个序列... 如:0 1 1 2 2 3 3 4 原序列为1 ...

  9. Java自学资料——线程

    [转]传智播客成都java培训中心学员笔记. 线程: static int MAX_PRIORITY 线程能够具有的最高优先级. static int MIN_PRIORITY 线程能够具有的最低优先 ...

  10. 为什么推荐std::string而不是char*

    例如如下: map<const char*, const char*> map_test; map_test["a"] = "a"; map_tes ...