POJ2155 Matrix 二维树状数组应用

一个N*N(1<=N<=1000)的矩阵,从(1,1)开始,给定一些操作C和一些查询Q,一共K条(1<=K<=50000):

C x1,y1,x2,y2 表示从x1行y1列到x2行y2列的元素全部反转(0变成1,1变成0);

Q x y表示询问x行y列的元素是0还是1。

题目乍一看感觉还是很难,如果能记录每一个元素的状态值,那答案是显而易见的,但是元素过多,如果每次都对每一个元素进行更新状态的话,复杂度太高。实际上只要记录边界的特定坐标的反转次数,最好的选择那就是二维树状数组了。

对于操作C,为了保证元素在子矩阵内的反转次数都增加了1,需要在Bit(x1,y1)的位置上加上1,Bit(x2 + 1, y1) 减去1,Bit(x1, y2 + 1)减去1,Bit(x2 +1, y2 + 1)加上1;

对于操作Q,直接获取Bit(x,y) 的值就代表该元素反转了多少次。

#include <stdlib.h>
#include <stdio.h>
#include <vector>
#include <math.h>
#include <string.h>
#include <string>
#include <iostream>

#define MAXN 1001

template <class TYPE>
void BITAdd(TYPE array[MAXN], int i, TYPE addvalue, int n)
{
	while (i <= n)
	{
		array[i] += addvalue;
		i += i & -i;
	}
}

template <class TYPE>
TYPE BITGet(TYPE array[MAXN], int i)
{
	TYPE ss = 0;
	while (i > 0)
	{
		ss += array[i];
		i -= i & -i;
	}
	return ss;
}

template <class TYPE>
void BIT2DAdd(TYPE array[MAXN][MAXN], int i, int j,TYPE addvalue, int n, int m)
{
	while (i <= n)
	{
		BITAdd<TYPE>(array[i], j, addvalue, m);
		i += i & -i;
	}
}

template <class TYPE>
TYPE BIT2DGet(TYPE array[MAXN][MAXN], int i, int j)
{
	TYPE ss = 0;
	while (i > 0)
	{
		ss += BITGet<TYPE>(array[i], j);
		i -= i & -i;
	}
	return ss;
}

typedef int MYTYPE;

MYTYPE Bit[MAXN][MAXN];

int main()
{
#ifdef _DEBUG
	freopen("d:\\in.txt", "r", stdin);
#endif
	int x;
	scanf("%d", &x);
	for (int i = 0; i < x; i++)
	{
		memset(Bit, 0, sizeof(Bit));
		int n, k;
		scanf("%d %d\n", &n, &k);
		if (i >= 1)
		{
			printf("\n");
		}
		for (int j = 0; j < k; j++)
		{
			char ope;
			scanf("%c", &ope);
			if (ope == 'C')
			{
				int x1, x2, y1, y2;

				scanf("%d %d %d %d\n", &x1, &y1, &x2, &y2);
				BIT2DAdd<int>(Bit, x1, y1, 1, n, n);
				BIT2DAdd<int>(Bit, x2 + 1, y1, -1, n, n);
				BIT2DAdd<int>(Bit, x1, y2 + 1, -1, n, n);
				BIT2DAdd<int>(Bit, x2 + 1, y2 + 1, 1, n, n);
			}
			else
			{
				int x, y;
				scanf("%d %d\n", &x, &y);

				int thisvalue = BIT2DGet<int>(Bit, x, y);
				printf("%d\n", thisvalue & 1);
			}
		}
	}
	return 0;
}
时间: 2024-10-13 19:42:26

POJ2155 Matrix 二维树状数组应用的相关文章

POJ2155 Matrix 二维树状数组的应用

有两种方法吧,一个是利用了树状数组的性质,很HDU1556有点类似,还有一种就是累加和然后看奇偶来判断答案 题意:给你一个n*n矩阵,然后q个操作,C代表把以(x1,y1)为左上角到以(x2,y2)为右下角的矩阵取反,意思就是矩阵只有0,1元素,是0的变1,是1的变0,Q代表当前(x,y)这个点的状况,是0还是1? 区间修改有点特别,但是若区间求和弄懂了应该马上就能懂得: add(x2,y2,1); add(x2,y1,-1);//上面多修改了不需要的一部分,所以修改回来 add(x1,y2,-

[poj2155]Matrix(二维树状数组)

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 25004   Accepted: 9261 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

POJ2155:Matrix(二维树状数组,经典)

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 upp

Matrix 二维树状数组的第二类应用

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 17976   Accepted: 6737 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

POJ 2155 Matrix(二维树状数组,绝对具体)

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 20599   Accepted: 7673 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

POJ 2155 Matrix(二维树状数组,绝对详细)

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 20599   Accepted: 7673 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

POJ 2155 Matrix 二维树状数组

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 19174   Accepted: 7207 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

POJ - 2155 Matrix (二维树状数组 + 区间改动 + 单点求值 或者 二维线段树 + 区间更新 + 单点求值)

POJ - 2155 Matrix Time Limit: 3000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Submit Status 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 ha

Matrix (二维树状数组)

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 upp