二维线段树

Posted 2021-08-31 一只特立独行的猫

线段树在用于一维空间中进行单点修改和区间查询非常具有优势，同样，在二维空间中线段树也可以实现该功能。可以查询区间和或者最小值。在面临多次单点修改和区间查询时，还是非常具有优势。如果需要进行区间修改和区间查询，那么就引入lazytag标记就可以实现优化。
对比一下一维和二维线段树划分区间的思路：
一维线段树划分区间思路：

二维线段树划分区间的思路：

大概就是上述思路,然后我们来看代码：

#include<iostream>

#define _1 node<<2		//第一象限
#define _2 node<<2|1	//第二象限
#define _3 node<<2|2	//第三象限
#define _4 node<<2|3	//第四象限
#define midx ((left + right) >> 1)
#define midy ((down + up) >> 1)

typedef long long ll;

using namespace std;

const int N = 3e2 + 5;

ll a[N][N],tree[(N<<4)*(N<<4)];

void treeUp(int node, int left, int right, int up, int down) {
	//向上更新
	//传入区间变量为了方便加lazytag
	tree[node] = tree[_1] + tree[_2] + tree[_3] + tree[_4];
	return;
}

void build(int node,int left,int right,int up,int down) {
	if (left > right || up > down) {
		return;
	}
	if (left == right && up == down) {
		//定位到一个格子上
		tree[node] = a[left][up];
		//cout << left << " " << up << " "<<tree[node]<<endl;
		return;
	}
	//cout << left << " " << right << " " << up << " " << down << endl;
	build(_1, midx + 1, right, up, midy);
	build(_2, left, midx, up, midy);
	build(_3, left, midx, midy + 1, down);
	build(_4, midx + 1, right, midy + 1, down);
	treeUp(node, left, right, up, down);
	return;
}

void updata(int node, int left, int right, int up, int down, int x, int y, int val) {
	//将x,y处的值更新为val
	if (x > right || x<left || y<up || y > down) {
		return;
	}
	if (left == right && up == down) {
		a[x][y] = val;
		tree[node] = val;
		return;
	}
	updata(_1, midx + 1, right, up, midy, x, y, val);
	updata(_2, left, midx, up, midy, x, y, val);
	updata(_3, left, midx, midy + 1, down, x, y, val);
	updata(_4, midx + 1, right, midy + 1, down, x, y, val);
	treeUp(node, left, right, up, down);
}

ll query(int node, int left, int right, int up, int down, int L, int R, int U, int D) {
	//查询LRUD围成的矩形的和
	if (L > right || R<left || D<up || U > down) {
		return 0;
	}
	if (left>=L&&R>=right&&up>=U&&down<=D) {
		return tree[node];
	}
	ll sum1 = query(_1, midx + 1, right, up, midy, L, R, U, D);
	ll sum2 = query(_2, left, midx, up, midy, L, R, U, D);
	ll sum3 = query(_3, left, midx, midy + 1, down, L, R, U, D);
	ll sum4 = query(_4, midx + 1, right, midy + 1, down, L, R, U, D);
	return sum1 + sum2 + sum3 + sum4;
}

int main() {
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			cin >> a[i][j];
		}
	}
	build(1, 1, m, 1, n);
	int k;
	cin >> k;
	while (k--) {
		int flag;
		cin >> flag;
		if (flag&1) {
			//更新操作
			int x, y, c;
			cin >> x >> y >> c;
			updata(1, 1, m, 1, n, x, y, c);
		}
		else {
			//查询操作
			int cnt = 0;
			int L, R, U, D;
			cin >> L >> R >> U >> D;
			cout << query(1, 1, m, 1, n, L,R,U,D) << endl;
		}
	}
	return 0;
}