二维线段树

2019-07-30 17:01:00 阅读：370 来源： 互联网

用于矩阵的快速修改和查询（修改和查询的时间复杂度都为logn，空间复杂度是（4*n）*（4*n））

这里提供两种操作单点修改+区间询查区间修改+单点询查

由于（单点修改+区间询查）和（区间修改+单点询查）的sum数组意义不同，所以该代码无法实现（单点修改+单点询查）和（区间修改+区间询查）

单点修改

void changex(int kx,int l,int r)
{
    changey(kx,1,1,h);
    if(l==r) return;
    int mid=l+r>>1;
    if(x<=mid) changex(kx<<1,l,mid);
    else changex(kx<<1|1,mid+1,r);
}
void changey(int kx,int ky,int l,int r)
{
    sum[kx][ky]++;
    if(l==r) return;
    int mid=l+r>>1;
    if(y<=mid) changey(kx,ky<<1,l,mid);
    else changey(kx,ky<<1|1,mid+1,r);
}

区间询查

void queryx(int kx,int l,int r)
{
    if(l>=xl && r<=xr)
    {
        queryy(kx,1,1,h);
        return;
    }
    int mid=l+r>>1;
    if(xl<=mid) queryx(kx<<1,l,mid);
    if(xr>mid) queryx(kx<<1|1,mid+1,r);
}
void queryy(int kx,int ky,int l,int r)
{
    if(l>=yl && r<=yr)
    {
        cnt+=sum[kx][ky];
        return;
    }
    int mid=l+r>>1;
    if(yl<=mid) queryy(kx,ky<<1,l,mid);
    if(yr>mid) queryy(kx,ky<<1|1,mid+1,r);
}

区间修改

void changex(int kx,int l,int r)
{
    if(x1<=l&&r<=x2)
    {
        changey(kx,1,1,n);
        return;
    }
    int mid=l+r>>1;
    if(x1<=mid) changex(kx<<1,l,mid);
    if(x2>mid) changex(kx<<1|1,mid+1,r);
}
void changey(int kx,int ky,int l,int r)
{
    if(y1<=l&&r<=y2)
    {
        sum[kx][ky]++;
        return;
    }
    int mid=l+r>>1;
    if(y1<=mid) changey(kx,ky<<1,l,mid);
    if(y2>mid) changey(kx,ky<<1|1,mid+1,r);
}

单点询查

void queryx(int kx,int l,int r)
{
    queryy(kx,1,1,n);
    if(l==r) return;
    int mid=l+r>>1;
    if(x<=mid) queryx(kx<<1,l,mid);
    else queryx(kx<<1|1,mid+1,r);
}
void queryy(int kx,int ky,int l,int r)
{
    ans+=sum[kx][ky];
    if(l==r) return;
    int mid=ly+ry>>1;
    if(y<=mid) queryy(kx,ky<<1,l,mid);
    else queryy(kx,ky<<1|1,mid+1,r);
}

Matrix POJ - 2155 （二维线段树区间修改+区间询查）

https://vjudge.net/contest/301590#problem/D

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

#include<stdio.h>
#include<string.h>
const int num=1005;
int tree[num*4][num*4];
int x1,y1,x2,y2,n,ans;
int pointx,pointy;
void changey(int kx,int ky,int treel,int treer)
{
    if(y1<=treel&&y2>=treer){
        tree[kx][ky]++;
        return;
    }
    int bet=(treel+treer)/2;
    if(y1<=bet) changey(kx,ky*2,treel,bet);
    if(y2>bet) changey(kx,ky*2+1,bet+1,treer);
}
void changex(int kx,int treel,int treer)
{
    if(x1<=treel&&x2>=treer){
        changey(kx,1,1,n);
        return;
    }
    int bet=(treel+treer)/2;    
    if(x1<=bet) changex(kx*2,treel,bet);
    if(x2>bet) changex(kx*2+1,bet+1,treer);
}
void asky(int kx,int ky,int treel,int treer)
{
    ans+=tree[kx][ky];
    if(treel==treer)
        return;
    int bet=(treel+treer)/2;
    if(pointy<=bet) asky(kx,ky*2,treel,bet);
    else asky(kx,ky*2+1,bet+1,treer);
}
void askx(int kx,int treel,int treer)
{
    asky(kx,1,1,n);
    if(treel==treer)
        return;
    int bet=(treel+treer)/2;
    if(pointx<=bet) askx(kx*2,treel,bet);
    else askx(kx*2+1,bet+1,treer);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        memset(tree,0,sizeof(tree));
        int m;
        scanf("%d %d",&n,&m);
        for(int i=0;i<m;i++){
            char ss[2];
            scanf("%s",ss);
            if(ss[0]=='C'){
                scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
                changex(1,1,n);
            }
            else{
                scanf("%d %d",&pointx,&pointy);
                ans=0;
                askx(1,1,n);
                printf("%d\n",ans%2);
            }
        }
        if(t!=0)
            printf("\n");
    }
    return 0;
}

标签：询查,单点,int,线段,treer,二维,kx,void
来源： https://blog.csdn.net/Chter0/article/details/90518674

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二维线段树