题解【POJ2155】Matrix
Description
Given an \(N \times N\) 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 \(A[i, j] = 0 (1 \leq i, j \leq N)\).
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.
C x1 y1 x2 y2
\((1 \leq x1 \leq x2 \leq n, 1 \leq y1 \leq y2 \leq n)\) changes the matrix by using the rectangle whose upper-left corner is \((x1, y1)\) and lower-right corner is \((x2, y2)\).Q x y
\((1 \leq x, y \leq n)\) querys \(A[x, y]\).
Input
The first line of the input is an integer \(X (X \leq 10)\) 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 \(T (2 \leq N \leq 1000, 1 \leq T \leq 50000)\) 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
Solution
题意简述:一个\(N \times N\)的\(01\)矩阵,和几种动态操作,包括对子矩阵\((x,y,xx,yy)\)的所有元素异或,查询某一点\((x,y)\)的元素值。
二维树状数组裸题。
异或的操作不难修改。
二维树状数组与一维树状数组不同的是:
- 一维树状数组维护的是一条链,而二维树状数组维护的却是一片区域。
- 一维树状数组更新和查找只有一重循环,而二维树状数组需要两重循环。
- 二维树状数组比一维树状数组多一维度。
经过以上分析,\(AC\)代码就不难得出了。
Code
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
int n, m, c[1003][1003], t;
bool fl = false;
inline void add(int x, int y)//二维树状数组的修改操作
{
for (int i = x; i <= n; i = i + (i & (-i)))
{
for (int j = y; j <= n; j = j + (j & (-j)))//注意两重循环
{
++c[i][j];
}
}
}
inline int getans(int x, int y)//二维树状数组的查询操作
{
int ans = 0;
for (int i = x; i; i = i - (i & (-i)))
{
for (int j = y; j; j = j - (j & (-j)))
{
ans = ans + c[i][j];//加上答案
}
}
return ans;
}
int main()
{
int t = gi();
while (t--)
{
if (!fl) fl = true;
else puts("");
n = gi(), m = gi();
memset(c, 0, sizeof(c));
while (m--)
{
char c;
cin >> c;
if (c == 'C')
{
int x = gi(), y = gi(), xx = gi(), yy = gi();
add(x, y), add(xx + 1, y), add(x, yy + 1), add(xx + 1, yy + 1);//进行插入操作
}
else
{
int x = gi(), y = gi();
printf("%d\n", getans(x, y) % 2);//输出最终答案
}
}
}
return 0;//结束
}
题解【POJ2155】Matrix的更多相关文章
- [LeetCode 题解] Spiral Matrix
前言 [LeetCode 题解]系列传送门: http://www.cnblogs.com/double-win/category/573499.html 题目链接 54. Spiral Matrix ...
- POJ2155 Matrix 【二维线段树】
题目链接 POJ2155 题解 二维线段树水题,蒟蒻本想拿来养生一下 数据结构真的是有毒啊,, TM这题卡常 动态开点线段树会TLE[也不知道为什么] 直接开个二维数组反倒能过 #include< ...
- POJ2155 Matrix 【二维树状数组】+【段更新点查询】
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 17766 Accepted: 6674 Descripti ...
- [poj2155]Matrix(二维树状数组)
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 25004 Accepted: 9261 Descripti ...
- [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 ...
- POJ2155:Matrix(二维树状数组,经典)
Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the ...
- C++题解:Matrix Power Series ——矩阵套矩阵的矩阵加速
Matrix Power Series r时间限制: 1 Sec 内存限制: 512 MB 题目描述 给定矩阵A,求矩阵S=A^1+A^2+--+A^k,输出矩阵,S矩阵中每个元都要模m. 数据范围: ...
- poj----2155 Matrix(二维树状数组第二类)
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 16950 Accepted: 6369 Descripti ...
- 【题解】Matrix BZOJ 4128 矩阵求逆 离散对数 大步小步算法
传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=4128 大水题一道 使用大步小步算法,把数字的运算换成矩阵的运算就好了 矩阵求逆?这么基础的线 ...
- 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 ...
随机推荐
- Linux系统下的CPU、内存、IO、网络的压力测试
本文转载自:小豆芽博客 一.对CPU进行简单测试: 1.通过bc命令计算特别函数 例:计算圆周率 echo "scale=5000; 4*a(1)" | bc -l -q MATH ...
- yamlapi接口测试框架
1.思路: yamlapi支持unittest与pytest两种运行模式, yamlapi即为yaml文件+api测试的缩写, 可以看作是一个脚手架工具, 可以快速生成项目的各个目录与文件, 只需维护 ...
- 使用Python爬虫整理小说网资源-自学
第一次接触python,原本C语言的习惯使得我还不是很适应python的语法风格.希望读者能够给出建议. 相关的入门指导来自以下的网址:https://blog.csdn.net/c406495762 ...
- bugku_web_变量1(CTF)
这道题考察php全局变量GLOBALS的用法,同样是个php审计题. 看一下代码: flag In the variable ! <?php error_reporting(0); includ ...
- Docker构建镜像过于缓慢解决-----Docker构建服务之部署和备份jekyll网站
参考原文链接:https://www.jianshu.com/p/e6b7e68f2ba7 来自<第一本Docker书>,我觉得很有趣,就记录一下 准备国内ubuntu镜像 每次构建Ubu ...
- 7-8 矩阵A乘以B (15分)
7-8 矩阵A乘以B (15分) 给定两个矩阵A和B,要求你计算它们的乘积矩阵AB.需要注意的是,只有规模匹配的矩阵才可以相乘.即若A有Ra行.Ca列,B有Rb行.Cb列, ...
- py 二级习题(turtle)
用turtle画一个正方形 import turtle turtle.penup() turtle.goto(-100,-100) turtle.pendown() turtle.begin_fill ...
- element-ui的upload组件的clearFiles方法
<template> <div> <el-button @click="clearFiles">重新上传</el-button> & ...
- 诡异的Integer
先看下面的这个代码,为什么同样的都是赋值,却得不到同样的结果,也没有超出int的范围啊?这是为什么? package ppt_test; public class StrangeIntegerBeha ...
- Xamarin.Android 开发,生成时提示“Resource.Drawable”未包含“BG”的定义
Xamarin Android提示Resource.Drawable”未包含“BG”的定义错误信息:error CS0117: '“Resource.Drawable”未包含“BG”的定义Xamari ...