POJ2155 Matrix(树状数组)

本文介绍了一种使用二维树状数组处理矩阵中特定区域元素翻转的算法。通过实现区间求和和单点修改操作,该方法能够高效地应对矩阵内元素的状态改变及查询请求。

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给出一个矩阵,成块的修改一些值,原来是1变0,原来是0变1,询问某个点最后的值。

二维树状数组,再利用区间求和单点修改实现,最后只记录修改的次数,结果对2取余即为答案。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N=1005;
int tree[N][N];
int n;

int lowbit(int i)
{
    return i&(-i);
}
void updata(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]++;
}
int query(int x,int y)
{
    int ans=0;
    for(int i=x;i>0;i-=lowbit(i))
        for(int j=y;j>0;j-=lowbit(j))
            ans+=tree[i][j];
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(tree,0,sizeof(tree));
        int q;
        scanf("%d %d",&n,&q);
        while(q--)
        {
            char sym;
            scanf(" %c ",&sym);
            if(sym=='C')
            {
                int x1,y1,x2,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                updata(x1,y1);
                updata(x1,y2+1);
                updata(x2+1,y2+1);
                updata(x2+1,y1);
            }
            if(sym=='Q')
            {
                int x,y;
                scanf("%d%d",&x,&y);
                int ans=query(x,y);
                printf("%d\n",ans%2);
            }
        }
        printf("\n");
    }
    return 0;
}

Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 31005 Accepted: 11323

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

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