• poj2155--Matrix(二维树状数组)


    Matrix
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 18021   Accepted: 6755

    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
    二维的树状数组。一般的改动和查询
    对二维的树状数组的改动
    void add(int i,int j,int d,int n)
    {
        int x , y ;
        for(x = i ; x <= n ; x += lowbit(x))
            for(y = j ; y <= n ; y += lowbit(y))
                c[x][y] += d;
    }
    查询
    int sum(int i,int j)
    {
        int a = 0 , x , y ;
        for(x = i ; x >= 0 ; x -= lowbit(x))
            for(y = j ; y >= 0 ; y -= lowbit(y))
                a += c[x][y] ;
        return a ;
    }
    题意:给出n*n的矩阵,矩阵中的值仅仅能为0或者1,初始值所有是0。C x1 y1 x2 y2 表示在(x1,y1)和(x2,y2)围成的矩形中的所有值变换 ;Q x y 查询该点当前的值。也能够觉得是统计该点经过了几次的变化。
    在这个题中树状数组由后向前更新,每个点的值表示由该点向前的全部点变换的次数
    对于C的操作
    首先 将(1,1)到(x2,y2)的全部点+1 ,代表有该点向前的矩形中的变化次数均+1。再由 (x1-1,y2)(x2,y1-1)均-1,(x1-1,y1-1)+1,平衡掉多操作的点,那么当计算该点变换次数时。由该点向后累加,得到的总和就是该点的变换次数。假设是奇数代表1,否则是0.

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    int c[1010][1010] ;
    int lowbit(int x)
    {
        return x & -x ;
    }
    void add(int i,int j,int d)
    {
        int x , y ;
        for(x = i ; x > 0 ; x -= lowbit(x))
            for(y = j ; y > 0 ; y -= lowbit(y))
                c[x][y] += d;
    }
    int sum(int i,int j,int n)
    {
        int a = 0 , x , y ;
        for(x = i ; x <= n ; x += lowbit(x))
            for(y = j ; y <= n ; y += lowbit(y))
                a += c[x][y] ;
        return a ;
    }
    int main()
    {
        int t , tt , i , j , n , m , x1 , y1 , x2 , y2 ;
        char ch ;
        scanf("%d", &t);
        for(tt = 1 ; tt <= t ; tt++)
        {
            scanf("%d %d", &n, &m);
            memset(c,0,sizeof(c));
            while(m--)
            {
                getchar();
                scanf("%c", &ch);
                if( ch == 'C' )
                {
                    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
                    add(x2,y2,1);
                    add(x1-1,y2,-1);
                    add(x2,y1-1,-1);
                    add(x1-1,y1-1,1);
                }
                else
                {
                    scanf("%d %d", &x1, &y1);
                    printf("%d
    ", sum(x1,y1,n)%2 );
    
                }
            }
            if(tt != t)
                printf("
    ");
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/wzjhoutai/p/7015706.html
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