Matrix+POJ+二维树状数组初步

Matrix

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

1

0

0

1

 
# include<iostream>

# include<cstdio>

# include<cstring>

# include<string>

# include<set>

# include<map>

# include<sstream>

using namespace std;

const int maxn=1000+5;

int tree[maxn][maxn];

char a[10];

int n;

int lowbit(int x)

{

    return x&-x;

}

int getsum(int x,int y)

{

    int s=0;

    for(int i=x;i>0;i-=lowbit(i))

        for(int j=y;j>0;j-=lowbit(j))

           s+=tree[i][j];

           if(s%2==0)  return 0;

            else

                return 1;

}

void update(int x,int y)

{

    for(int i=x;i<=n;i+=lowbit(i))

        for(int j=y;j<=n;j+=lowbit(j))

            tree[i][j]^=1;

}

int main()

{

    /*#ifndef ONLINE_JUDGE

       freopen("in.cpp","r",stdin);

       freopen("out.cpp","w",stdout);

    #endif*/

    int t;

    scanf("%d",&t);

    while(t--)

    {

        int m;

        scanf("%d%d",&n,&m);

        for(int i=1; i<=n; i++)

            for(int j=1; j<=n; j++)

               tree[i][j]=0;

        while(m--)

        {

            int x1,y1,x2,y2;

           scanf("%s%d%d",&a[0],&x1,&y1);

           if(a[0]=='C')

           {

               scanf("%d%d",&x2,&y2);

               update(x2+1,y2+1);

               update(x2+1,y1);

               update(x1,y2+1);

               update(x1,y1);

           }

           else if(a[0]=='Q')

           {

               printf("%d\n",getsum(x1,y1));

           }

        }

        cout<<endl;

    }

    return 0;

}

巴特西

Matrix+POJ+二维树状数组初步

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