POJ 2155 Matrix（二维树状数组）

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. <br> <br>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. <br>

 

Output

For each querying output one line, which has an integer representing A[x, y]. <br> <br>There is a blank line between every two continuous test cases. <br>

 

Sample Input

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

 

Sample Output

1001

基本的二维数组数组，这里运用的是区间修改，单点求和，

类似题目：基本二维树状数组

不懂看文章：https://wenku.baidu.com/view/1e51750abb68a98271fefaa8

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>#include<vector>using namespace std;typedef long long ll;#define M 1005int n,m;int tree[M][M];inline int lowbit(int i){    return i&(-i);}void add(int x,int y,int v){    int i,j;    for(i=x;i<=n;i+=lowbit(i))        for(j=y;j<=n;j+=lowbit(j))    {        tree[i][j]+=v;    }}int sum(int x,int y){    int i,j,res=0;    for(i=x;i>0;i-=lowbit(i))        for(j=y;j>0;j-=lowbit(j))    {        res+=tree[i][j];    }    return res;}int main(){    int T,x1,x2,y1,y2;    char ch;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        memset(tree,0,sizeof(tree));        while(m--)        {            getchar();            ch=getchar();            if(ch=='C')            {                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                add(x1,y1,1);                add(x2+1,y2+1,1);                add(x1,y2+1,1);                add(x2+1,y1,1);            }            else            {                scanf("%d%d",&x1,&y1);                printf("%d\n",sum(x1,y1)&1);            }        }        if(T) printf("\n");    }    return 0;}