POJ - 2155 Matrix —— 二维树状数组

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 29525		Accepted: 10783

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

题意：给定一个矩阵 两个操作 一个是给出两个坐标 将坐标所形成的矩阵中的所有0变为1 1变为0 另一个是询问某个点是多少

思路：二维的树状数组

①

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#define max_ 1010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;

int n,m;
int c[max_][max_];
int lb(int x)
{
	return (x&(-x));
}
void update(int x,int y,int v)
{
	for(int i=x;i<=n;i+=lb(i))
		for(int j=y;j<=n;j+=lb(j))
			c[i][j]+=v;
}
int getsum(int x,int y)
{
	int sum=0;
	for(int i=x;i>0;i-=lb(i))
		for(int j=y;j>0;j-=lb(j))
			sum+=c[i][j];
	return sum;
}
int main(int argc, char const *argv[])
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		memset(c,0,sizeof(c));
		char c;
		int x1,x2,y1,y2;
		while(m--)
		{
			scanf(" %c",&c);
			if(c=='C')
			{
				scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
				update(x2+1,y2+1,1);
				update(x1,y1,1);
				update(x1,y2+1,-1);
				update(x2+1,y1,-1);
			}
			else
			{
				scanf("%d%d",&x1,&y1);
				printf("%d\n",getsum(x1,y1)&1);
			}
		}
		printf("\n");
	}
	return 0;
}

②

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#define max_ 1010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int n,m;
int c[max_][max_];
int lb(int x)
{
	return (x&(-x));
}
void update(int x,int y)
{
	for(int i=x;i<=n;i+=lb(i))
		for(int j=y;j<=n;j+=lb(j))
			c[i][j]++;
}
int getsum(int x,int y)
{
	int sum=0;
	for(int i=x;i>0;i-=lb(i))
		for(int j=y;j>0;j-=lb(j))
			sum+=c[i][j];
	return sum;
}
int main(int argc, char const *argv[])
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(c,0,sizeof(c));
		scanf("%d%d",&n,&m);
		char c;
		int x1,x2,y1,y2;
		while(m--)
		{
			scanf(" %c",&c);
			if(c=='C')
			{
				scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
				x1++;x2++;y1++;y2++;
				update(x2,y2);
				update(x1-1,y2);
				update(x2,y1-1);
				update(x1-1,y1-1);
			}
			else if(c=='Q')
			{
				scanf("%d%d",&x1,&y1);
				printf("%d\n",getsum(x1,y1)&1);
			}
		}
		printf("\n");
	}	
	return 0;
}