二维树状数组

Matrix

二维树状数组入门题

poj 2155

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题目描述：

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

题目的意思就是说，让你更新一个区域内的值，每次更新，矩阵里面的值就取反一下，就是1变0，0变1；然后问你一个点，问他的状态。

一开始，我想的是，我只学了一维的树状数组，我就把它转化成一个n*n的一维线段树，然后焦一发，然后t了；

上网看题解，发现，这题要用到二维的树状数组来写，然后就写偏博客记录下这个二维树状数组是干啥的；怎么操作的，输进去什么得到什么；

就比如 我现在给你（2，2）（3，3）让你更新
我找的这个模版是这样解的
我先确定一个右下标，我通过四次操作来解决这个问题，讲不清楚，看例子吧

然后就是这题的代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<queue>#include<stack>#include<vector>#include<cmath>#include<set>#include<cstdlib>#include<functional>#include<climits>#include<cctype>#include<iomanip>using namespace std;typedef long long ll;#define INF 0x3f3f3f3f#define mod 1e9+7#define clr(a,x) memset(a,x,sizeof(a))const double eps = 1e-6;// 知识点分解：二维树状数组const int MAXN = 1005;int treeNum[MAXN][MAXN];int lowbit(int x){    return x&(-x);}long long getSum(int x , int y){    long long sum = 0;    for(int i = x ; i > 0 ; i -= lowbit(i))        for(int j = y ; j > 0 ; j -= lowbit(j))            sum += treeNum[i][j];    return sum;}void add(int x , int y , int val){    for(int i = x ; i < MAXN ; i += lowbit(i))        for(int j = y ; j < MAXN ; j += lowbit(j))            treeNum[i][j] += val;}int main(){    int t;    cin>>t;    while(t--)    {        int n,q;        clr(treeNum,0);        scanf("%d%d",&n,&q);        while(q--)        {            char c;            getchar();            scanf("%c",&c);            if(c=='C')            {                int a,b,c,d;                scanf("%d%d%d%d",&a,&b,&c,&d);                add(a,b,1);                add(a,d+1,-1);                add(c+1,b,-1);                add(c+1,d+1,1);            }            else            {              int a,b;               scanf("%d%d",&a,&b);              ll k =getSum(a,b);              if(k%2==0 ) printf("0\n");              else printf("1\n");            }        }        cout<<endl;    }    return 0;}

               ---langman