Matrix (二维树状数组)

传送门：POJ - 2155

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.

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).

Q x y (1 <= x, y <= n) querys A[x, y].

输入

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.

输出

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.

样例

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

Output
1
0
0
1

题解

题意：
C——以(x1,y1)为左上角，(x2,y2)为右上角的矩阵，每个元素01取反
Q——询问(x,y)的值

结合矩阵翻转，将区间修改单点查询转变成单点修改区间求和。

用二维树状数组保存点的修改次数，修改的时候进行如下操作：

黄色部分为本次需要修改的区间，我们对四个绿色点进行add操作，也就是将每个绿色点右下角部分的矩阵进行翻转，这样我们进行四次单点修改就完成了整个区间的修改。而要查询某个点的值时，求出它的翻转次数即可

Code

#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#define LL long long int
#define INIT(a,b) memset(a,b,sizeof(a))
//b——0,-1,128,0x3f,127 ,字符 
const double Pi = acos(-1);
const double E = exp(1.0); 
const LL mod =1e9+7;
const int MAX=0x7fffffff;
const int MIN=-0x7fffffff;
const int INF=0x3f3f3f3f;
using namespace std;
int N,M,tree[1010][1010];
int lowbit(int x){
    return x & -x;
}
void add(int x,int y,int num){
    for(int i=x;i<=N;i+=lowbit(i))
        for(int j=y;j<=N;j+=lowbit(j))
            tree[i][j]+=num;
}
int count(int x,int y){
    int sum=0;
    for(int i=x;i>0;i-=lowbit(i))
        for(int j=y;j>0;j-=lowbit(j))
            sum+=tree[i][j];
    return sum;
}
int main()
{
    int _,x1,y1,x2,y2;
    scanf("%d",&_);
    while(_--){
        INIT(tree,0);
        scanf("%d%d",&N,&M);
        char x;
        while(M--){
            getchar();
            scanf("%c",&x);
            if(x=='C'){
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                add(x1,y1,1);
                add(x2+1,y1,1);
                add(x1,y2+1,1);
                add(x2+1,y2+1,1);
            }
            else if(x=='Q'){
                scanf("%d%d",&x1,&y1);
                printf("%d\n",count(x1,y1)%2);
            }			
        }
        if(M)printf("\n");
    }	
    return 0;
}