POJ P2155 Matrix

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本文介绍了一种利用二维树状数组解决特定矩阵操作问题的方法，通过执行单点修改和查询操作来实现矩阵的高效更新和访问。特别地，文章详细解释了如何在矩阵上应用not操作，并利用树状数组的特性进行优化。

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 18581		Accepted: 7001

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

当年楼教主出的题目，当时在NOIP时代流行的一句话是虽然我不会这道题，但是我能AC它。这道题就是典型的这种题。此题乍看之下区间修改加询问，可以用二维线段树做，不过我不会建议你这么试的，因为实在是太烦了，而且这题还卡空间，你还不一定能过。仔细一看会发现它是单点询问，而不是区间询问，是楼教主出题出简单了呢，还是另有隐情。答案是后者，这题用到了树状数组神奇的性质，本来树状数组是用来区间查询，单点修改，这题刚好倒了一下，即用二维树状数组区间查询的方法来修改区间，而用单点修改的方法来单点查询，真的是神奇，我说不上为什么这么做可以，但是却可以AC这道题。

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<string>
#include<cstring>
#include<algorithm>
#include<fstream>
#include<queue>
#include<stack> 
#include<vector>
#include<cmath>
#include<iomanip>
#define rep(i,n) for(i=1;i<=n;i++)
#define MM(a,t) memset(a,t,sizeof(a))
#define lowbit(x) x&(-x)
#define fan(a) 1-a 
#define INF 1e9
typedef long long ll;
#define mod 1000000007
using namespace std;
int n,m;
int bita[1020][1020];
void update(int x,int y){
  int i,j;
  
  for(i=x;i>0;i-=lowbit(i))
    for(j=y;j>0;j-=lowbit(j))
      bita[i][j]=fan(bita[i][j]);
  
}
int query(int x,int y){
	int i,j,res=0;
	
	for(i=x;i<=n;i+=lowbit(i))
	  for(j=y;j<=n;j+=lowbit(j)){
          if(bita[i][j]) res=fan(res);
  	    }
   
   return res;
}
int main()
{
	int i,j,T;
	
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&n,&m);
		MM(bita,0);
		rep(i,m){
			char op[10];
			scanf("%s",op);
			if(op[0]=='C'){
				int x1,y1,x2,y2;
				scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
				update(x1-1,y1-1); update(x2,y2);
				update(x1-1,y2); update(x2,y1-1);
			}
			else{
			  int x,y;
			  scanf("%d%d",&x,&y);
			  printf("%d\n",query(x,y));	
			}
		}
		if(T!=0) cout<<'\n';
	}
	
	return 0;
}