• poj-2155 Matrix(二维树状数组)


    题目链接:

    Matrix

    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 23170   Accepted: 8613

    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
    AC代码:
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    const int N=1e3+4;
    int sum[N][N],n,l,r,xl,xr,yl,yr,m;
    char c;
    int lowbit(int x)
    {
        return x&(-x);
    }
    void update(int x,int y,int num)
    {
    
        while(y<=n)
        {
            int fx=x;
            while(fx<=n)
            {
                sum[fx][y]+=num;
                fx+=lowbit(fx);
            }
            y+=lowbit(y);
        }
    }
    int query(int x,int y)
    {
        int s=0;
        while(y>0){
                int fx=x;
        while(fx>0)
        {
            s+=sum[fx][y];
            fx-=lowbit(fx);
        }
        y-=lowbit(y);
        }
        return s;
    }
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            memset(sum,0,sizeof(sum));
            scanf("%d%d",&n,&m);
            for(int i=0;i<m;i++)
            {
                cin>>c;
                if(c=='C')
                {
                    scanf("%d%d%d%d",&xl,&yl,&xr,&yr);
                    update(xl,yl,1);
                    update(xl,yr+1,1);
                    update(xr+1,yl,1);
                    update(xr+1,yr+1,1);
                }
                else
                {
                    scanf("%d%d",&l,&r);
                    printf("%d
    ",query(l,r)%2);
                }
            }
            printf("
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5317431.html
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