【poj2155】【Matrix】二位树状数组
[pixiv] https://www.pixiv.net/member_illust.php?mode=medium&illust_id=34310873
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
题目大意
给一个N*N的矩阵,里面的值不是0,就是1。初始时每一个格子的值为0。
现对该矩阵有两种操作:(共T次)
1.C x1 y1 x2 y2:将左上角为(x1, y1),右下角为(x2, y2)这个范围的子矩阵里的值全部取反。
2.Q x y:查询矩阵中第i行,第j列的值。
根据数据范围,横纵两个方向都必须是log级的复杂度。如果按照题目原意直接模拟,区间修改单点查询,需要用线段树。但是我并不会二位线段树。那么就利用差分的思想,使其转化为单点修改区间查询,可以用树状数组来维护。
一维的差分是这样的
[le,ri]+val
那么二维的就是
但是详细的,(x,y)+1,是指的从(x,y)到(n,n)的矩阵都+1
那么根据容斥原理
由此一来,单点查询时就查询(0,0)到(x,y)的和
现在就是二维树状数组怎么实现的问题了
其实很简单,就是两个for套在一起就是了
void modify(int x,int y,int val){
for(int i=x;i<=n;i+=(i&(-i)))
for(int j=y;j<=n;j+=(j&(-j)))
c[i][j]++;
}
完整代码
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1000+5;
const int T=50000+5;
int n,t;
int c[N][N];
void modify(int x,int y,int val){
for(int i=x;i<=n;i+=(i&(-i)))
for(int j=y;j<=n;j+=(j&(-j)))
c[i][j]++;
}
int query(int x,int y){
int rt=0;
for(int i=x;i>0;i-=(i&(-i)))
for(int j=y;j>0;j-=(j&(-j)))
rt+=c[i][j];
return rt;
}
void solve(){
scanf("%d%d",&n,&t);
memset(c,0,sizeof(c));
while(t--){
char opt[2];
scanf("%s",opt);
if(opt[0]=='C'){
int x1,y1,x2,y2;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
modify(x1,y1,1);
modify(x1,y2+1,-1);
modify(x2+1,y1,-1);
modify(x2+1,y2+1,1);
}
else{
int x,y;
scanf("%d%d",&x,&y);
printf("%d\n",query(x,y)%2);
}
}
printf("\n");
}
int main(){
int x;
scanf("%d",&x);
while(x--) solve();
return 0;
}
总结:
1、看到操作不必直接模拟,用差分等思想可以化难为简
【poj2155】【Matrix】二位树状数组的更多相关文章
- [poj2155]Matrix(二维树状数组)
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 25004 Accepted: 9261 Descripti ...
- 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 ...
- 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 ...
- 【poj2155】Matrix(二维树状数组区间更新+单点查询)
Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the ...
- poj2155一个二维树状数组
...
- POJ 2155 Matrix(二维树状数组,绝对具体)
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 20599 Accepted: 7673 Descripti ...
- POJ 2155:Matrix 二维树状数组
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 21757 Accepted: 8141 Descripti ...
- HDU4456-Crowd(坐标旋转+二位树状数组+离散化)
转自:http://blog.csdn.net/sdj222555/article/details/10828607 大意就是给出一个矩阵 初始每个位置上的值都为0 然后有两种操作 一种是更改某个位置 ...
- poj 2155 Matrix (二维树状数组)
题意:给你一个矩阵开始全是0,然后给你两种指令,第一种:C x1,y1,x2,y2 就是将左上角为x1,y1,右下角为x2,y2,的这个矩阵内的数字全部翻转,0变1,1变0 第二种:Q x1 y1,输 ...
随机推荐
- NGUI执行基本事件的原理
通常我们为对象附加一个脚本组件,脚本组件只要加此鼠标处理事件方法,这个对象就有了点击事件了: void OnClick() { Debug.Log("onclick"); } 可为 ...
- wxPython 安装 及参考文档
三种操作平台上的安装方法 1.windows 和 mac pip install -U wxPython 2.linux pip install -U -f https://extras.wxpyth ...
- ipa和ironic-conductor交互
IPA使用lookup和hearteat机制与Ironic Conductor进行交互,启动时agent给Conductor的vendor_passthru lookup endpoint(地址为/v ...
- 云效(阿里云)流水线 + nginx + uWsgi + flask + python3 基础环境搭建 --备忘
一.开发环境搭建 1.安装python3 yum -y groupinstall "Development tools" yum -y install zlib-devel bzi ...
- c++知识点总结-模板特化
类模板的全特化与偏特化 类模板 template<typename T1, typename T2> class Test { public: Test(T1 i,T2 j):a(i),b ...
- intellij idea 2017 工具使用问题
1.打开idea 打开maven项目报错:Unable to import maven project 2.在idea中Help->Show Log in Explorer->idea.l ...
- aes加密码
js地址 https://github.com/yves8888/crypto-js 下面src<!DOCTYPE html> <html lang="en"&g ...
- 【转】VS常用快捷键
每次在网上搜关于VS有哪些常用快捷键的时候,出来的永远是一串长的不能再长的列表,完全没体现出“常用”二字,每次看完前面几个就看不下去了,相信大家都 有这种感觉.其实我们平时用的真的只有很少的一部分,借 ...
- [codeforces] 527A Playing with Paper
原题 简单的gcd #include<cstdio> #include<algorithm> typedef long long ll; using namespace std ...
- Jquery不同版本共用的解决方案(插件编写)
最近在为某公司做企业内部UI库,经过研究分析和评审,决定基于Jquery开发,结合Bootstrap插件那简洁,优雅,高效的思想进行插件编写. 但是在编写的过程中遇到一个头疼的问题,就是正在编写的插件 ...