Matrix (poj 2155)

传送门

Time Limit: 3000MS

Memory Limit: 65536K

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
Source
POJ Monthly,Lou Tiancheng

题意

对于一个0,1矩阵，支持左下(x1,y1)-右上(x2,y2)矩形区域内的反转，然后对单点(x,y)的查询。

Solution

今天才学的二维树状数组（菜的一匹），其实看到对于一个二维区域的维护应该就能想到这类型的数据结构。那么对于单点查询，结合树状数组可想到是由差分来实现的，因为树状数组只能提取前缀和。然后对于一个二维平面如何加值，类似于一维中的s[x]+v,s[y]-v。推广到二维则有：

可以看一下这个图。

另外题面中的加大的字各位可以品尝..QAQ

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#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int s[2100][2100],n,q;
int lowbit(int x){return x&(-x);}
void add(int x,int y,int c)
{
	for(int i=x;i<=n;i+=lowbit(i))
		for(int j=y;j<=n;j+=lowbit(j))
			s[i][j]+=c;
}
int query(int x,int y)
{
	int rtn=0;
	for(int i=x;i;i-=lowbit(i))
		for(int j=y;j;j-=lowbit(j))
			rtn+=s[i][j];
	return rtn;	
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&q);
		memset(s,0,sizeof(s));
		while(q--)
		{
			char ch[5];
			scanf("%s",ch);
			if(ch[0]=='C')
			{
				int x1,y1,x2,y2;
				scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
				add(x2+1,y2+1,1);
				add(x1,y2+1,-1);
				add(x2+1,y1,-1);
				add(x1,y1,1);
			}
			if(ch[0]=='Q')
			{
				int x,y;
				scanf("%d%d",&x,&y);
				printf("%d\n",query(x,y)%2);
			}
		}
		printf("\n");
	}
	return 0;
} 
/*
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
*/

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