二维树状数组模版,唯一困难,看题!!(其实是我英语渣)

Matrix

Time Limit: 3000MS Memory Limit: 65536K

Total Submissions: 22098 Accepted: 8240

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].

    题目大意:

    给出t个n*n的矩阵,初始都是0,并给一个m,给出m个命令:

    命令“C x1 y1 x2 y2”将(x1,y1)–(x2,y2)上每个点进行交换(0变为1,1变为0)

    命令“Q x y”求(x,y)的值

    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

前排膜娄教主%%%

这个题的话,值得一提的就是:
在修改(x1,y1)--(x2,y2)的时候,应用区间修改的原理只需要修改(x1,y1),(x2+1,y1),(x1,y2+1),(x2+1,y2+1)即可
以及每个记录的值是变换过几次的值,所以结果%2即可
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int matrix[2000][2000]={0};
int t,n; int lowbit(int x)
{
return x&(-x);
} int sum(int x,int y)
{
int total=0;
for (int i=x; i>0; i-=lowbit(i))
for (int j=y; j>0; j-=lowbit(j))
total+=matrix[i][j];
return total;
} void change(int x,int y)
{
for (int i=x; i<=n; i+=lowbit(i))
for (int j=y; j<=n; j+=lowbit(j))
matrix[i][j]++;
} int main()
{
scanf("%d",&t);
for (int T=1; T<=t; T++)
{
int m;
scanf("%d%d",&n,&m);
memset(matrix,0,sizeof(matrix));
while (m>0)
{
char command[10];
scanf("%s",&command);
if (command[0]=='C')
{
int x1,x2,y1,y2;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
change(x1,y1);
change(x2+1,y1);
change(x1,y2+1);
change(x2+1,y2+1);
}
if (command[0]=='Q')
{
int x,y;
scanf("%d%d",&x,&y);
int ans=sum(x,y) % 2;
printf("%d\n",ans);
}
m--;
}
printf("\n");
}
return 0;
}

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