poj-2155 matrix :二维树状数组
2017-04-07 20:03
423 查看
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
Sample Output
Source
POJ Monthly,Lou Tiancheng
题意:有一个矩阵,里面所有数字都是0,有两种操作,c操作,把一个矩形里的数字都取非,q操作,展示某个点的值。
备注:基础一维http://www.hankcs.com/program/algorithm/poj-3109-inner-vertices.html
http://blog.csdn.net/libin56842/article/details/46620445 http://blog.csdn.net/int64ago/article/details/7429868:搞懂树状数组 http://download.csdn.net/detail/lenleaves/4548401:论文二进制思想的应用
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 26943 | Accepted: 9873 |
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
Source
POJ Monthly,Lou Tiancheng
题意:有一个矩阵,里面所有数字都是0,有两种操作,c操作,把一个矩形里的数字都取非,q操作,展示某个点的值。
#include <iostream> #include<string.h> #include<cstdio> #include<algorithm> #include<queue> #include<cstdlib> using namespace std; const int Max = 1128; int data[Max][Max],n,k; int lowbit(int x)//求出该移动的距离 { return x&-x; } void add(int x,int y,int w)//维护树状数组更新,从前往后; { for(int i=x;i<=n;i+=lowbit(i)) { for(int j=y;j<=n;j+=lowbit(j)) { data[i][j]+=w; } } } int sum(int x,int y) { int ans = 0; for(int i = x;i>0;i-=lowbit(i))//求出该点处更改次数,从该点向左一直到i<=0 { for(int j = y;j>0;j-=lowbit(j)) { ans += data[i][j]; } } return ans; } int main() { char str[103]; int t; cin>>t; while(t--) { cin>>n>>k; memset(data,0,sizeof(data));//第一组数据对,wr一般是未进行初始化,其中初始化时sizeof用数组名 while(k--) { scanf("%s",str);//有字符数字空格输入,推荐采用字符串输入 if(str[0]=='C') { int x1,y1,x2,y2; scanf("%d%d%d%d",&x1,&y1,&x2,&y2);//输入数据推荐使用scanf add(x1,y1,1); add(x1,y2+1,-1); add(x2+1,y1,-1); add(x2+1,y2+1,1);//树状数组跟新左侧不变,右侧加1,即(x,y+1) //此处用到了容斥原理 } else { int x,y; scanf("%d%d",&x,&y); cout<<sum(x,y)%2<<endl;//更改次数跟2取余即为翻转 } } if(t) cout<<endl;//看准题意,两组数据之间有一组空格 } return 0; }
备注:基础一维http://www.hankcs.com/program/algorithm/poj-3109-inner-vertices.html
POJ 3109 Inner Vertices
本题参考博客:http://blog.csdn.net/libin56842/article/details/46620445 http://blog.csdn.net/int64ago/article/details/7429868:搞懂树状数组 http://download.csdn.net/detail/lenleaves/4548401:论文二进制思想的应用
相关文章推荐
- poj 2155 matrix 二维树状数组反应用
- poj2155 Matrix 【二维树状数组】
- 【二维树状数组】poj 2155 Matrix
- POJ 2155 Matrix (二维树状数组,区间更新,点查找)
- 二维树状数组-POJ-2155-Matrix
- poj 2155- Matrix (树状数组,二维,更新区间,查询单点)
- poj2155 Matrix 二维树状数组
- poj 2155 Matrix 二维树状数组 (经典)
- POJ2155 Matrix 二维树状数组的应用
- 【二维树状数组--模板】poj 2155 Matrix、poj 1195 Mobile phones
- POJ 2155 Matrix ( 二维树状数组 ) || HDU 3584 Cube ( 三维树状数组 )
- POJ 2155 Matrix (二维线段树入门,成段更新,单点查询 / 二维树状数组,区间更新,单点查询)
- POJ 2155 Matrix 【二维树状数组】
- poj 2155 Matrix 二维树状数组
- poj 2155 Matrix 二维树状数组
- POJ-2155 Matrix (二维树状数组 入门题)
- POJ 2155 Matrix 二维树状数组
- POJ-2155 Matrix 二维树状数组
- POJ 2155 Matrix【二维树状数组】POJ 2155【
- POJ 2155 Matrix(二维树状数组,绝对具体)