POJ - 2155 Matrix —— 二维线段树

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本文介绍了一种解决矩阵区间更新及点查询问题的方法。通过建立四叉树数据结构，实现矩阵元素的高效批量反转操作及查询指定位置的当前状态。文章提供了完整的代码示例，包括初始化、更新和查询等关键步骤。

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Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 32486		Accepted: 11776

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

区间更新点查询

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath>
#include <vector>
#include <bitset>
#define max_ 1100
#define inf 0x3f3f3f3f
#define ll long long
#define les 1e-8
#define mod 364875103
using namespace std;
int tree[max_*4][max_*4];
int n,m;
void buil(int i,int l,int r,int id)
{
    tree[id][i]=0;
    if(l==r)
    return;
    int mid=(l+r)>>1;
    buil(i<<1,l,mid,id);
    buil(i<<1|1,mid+1,r,id);
}
void built(int i,int l,int r)
{
    buil(1,1,n,i);
    if(l==r)
    return;
    int mid=(l+r)>>1;
    built(i<<1,l,mid);
    built(i<<1|1,mid+1,r);
}
void upda(int i,int l,int r,int nl,int nr,int id)
{
    if(l==nl&&r==nr)
    {
        tree[id][i]++;
        tree[id][i]%=2;
        return;
    }
    int mid=(nl+nr)>>1;
    if(r<=mid)
    upda(i<<1,l,r,nl,mid,id);
    else if(l>mid)
    upda(i<<1|1,l,r,mid+1,nr,id);
    else
    {
        upda(i<<1,l,mid,nl,mid,id);
        upda(i<<1|1,mid+1,r,mid+1,nr,id);
    }
}
void updata(int i,int l,int r,int nl,int nr,int dl,int dr)
{
    if(l==nl&&r==nr)
    {
        upda(1,dl,dr,1,n,i);
        return;
    }
    int mid=(nl+nr)>>1;
    if(r<=mid)
    updata(i<<1,l,r,nl,mid,dl,dr);
    else if(l>mid)
    updata(i<<1|1,l,r,mid+1,nr,dl,dr);
    else
    {
        updata(i<<1,l,mid,nl,mid,dl,dr);
        updata(i<<1|1,mid+1,r,mid+1,nr,dl,dr);
    }
}
int que(int i,int x,int nl,int nr,int id)
{
    if(nl==nr)
    {
        return tree[id][i];
    }
    int mid=(nl+nr)>>1;
    if(x<=mid)
    return tree[id][i]%2+que(i<<1,x,nl,mid,id);
    else
    return tree[id][i]%2+que(i<<1|1,x,mid+1,nr,id);
}
int query(int i,int x,int y,int nl,int nr)
{
    if(nl==nr)
    return que(1,y,1,n,i);
    int mid=(nl+nr)>>1;
    if(x<=mid)
    return que(1,y,1,n,i)+query(i<<1,x,y,nl,mid);
    else
    return que(1,y,1,n,i)+query(i<<1|1,x,y,mid+1,nr);
}
int main(int argc, char const *argv[]) {
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        built(1,1,n);
        while(m--)
        {
            char c;
            scanf(" %c",&c);
            if(c=='C')
            {
                int x1,x2,y1,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                updata(1,x1,x2,1,n,y1,y2);
            }
            else
            {
                int x,y;
                scanf("%d%d",&x,&y);
                printf("%d\n",query(1,x,y,1,n)%2);
            }
        }
        if(t!=0)
        printf("\n" );
    }
    return 0;
}