二维线段树

最新推荐文章于 2022-02-03 02:43:47 发布

Scar_Halo

最新推荐文章于 2022-02-03 02:43:47 发布

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分类专栏： ACM - 树套树文章标签：二维线段树树套树矩阵和

本文链接：https://blog.youkuaiyun.com/Scar_Halo/article/details/104204406

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ACM - 树套树专栏收录该内容

4 篇文章

订阅专栏

维护一个矩阵，实现以下几个需求：
$① ：$ 单点修改，区间查最值
$② ：$ 区间加，区间查和
$③ ：$ 区间修改（递增），区间查最值

二维线段树本质上是树套树
因此无法实现 $p u s h d o w n$ ，用标记永久化替代
暂时无法满足任意模式的区间修改，也可能是我太菜了
注意修改后的 $m e r g e$ ，只针对所修改区间进行 $m e r g e$ 即可
参考：线段树套线段树
建树复杂度： $O(n^2)$
单次修改查询复杂度： $O(log^2n)$

例题一：Census UVA - 11297

建树后，单点修改，区间查最大最小值

// https://vjudge.net/problem/UVA-11297
#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
using pii = pair <int,int>;
const int maxn = 5e2 + 5;
int n, q;
ll a[maxn][maxn];
struct seg{
	ll t1[maxn<<2], t2[maxn<<2], lz[maxn<<2];
	void build(int L, int l, int r, int rt){
		if(l == r) {
			t1[rt] = t2[rt] = a[L][r];
			return;
		}
		int m = l + r >> 1;
		build(L, l, m, rt<<1);
		build(L, m+1, r, rt<<1|1);
		t1[rt] = max(t1[rt<<1], t1[rt<<1|1]);
		t2[rt] = min(t2[rt<<1], t2[rt<<1|1]);
	}
	void update(int L, int R, ll x, int l, int r, int rt){
		t1[rt] = t2[rt] = x;
		if(l==L && r==R){
			lz[rt] = x;
			return;
		}
		int m = l + r >> 1;
		if(m >= R) update(L, R, x, l, m, rt<<1);
		else if(L > m) update(L, R, x, m+1, r, rt<<1|1);
		else update(L, m, x, l, m, rt<<1), \
			 update(m+1, R, x, m+1, r, rt<<1|1);
		t1[rt] = max(t1[rt<<1], t1[rt<<1|1]);
		t2[rt] = min(t2[rt<<1], t2[rt<<1|1]);
	}
	ll query1(int L, int R, int l, int r, int rt, ll sum){
		if(l==L && r==R) return t1[rt];
//		sum += lz[rt];
		int m = l + r >> 1;
		if(m >= R) return query1(L, R, l, m, rt<<1, sum);
		else if(L > m) return query1(L, R, m+1, r, rt<<1|1, sum);
		else return max( query1(L, m, l, m, rt<<1, sum), \
						 query1(m+1, R, m+1, r, rt<<1|1, sum) );
	}
	ll query2(int L, int R, int l, int r, int rt, ll sum){
		if(l==L && r==R) return t2[rt];
//		sum += lz[rt];
		int m = l + r >> 1;
		if(m >= R) return query2(L, R, l, m, rt<<1, sum);
		else if(L > m) return query2(L, R, m+1, r, rt<<1|1, sum);
		else return min( query2(L, m, l, m, rt<<1, sum), \
						 query2(m+1, R, m+1, r, rt<<1|1, sum) );
	}
} t[maxn<<2], lz[maxn<<2]; 

void merge(seg &p, seg &p1, seg &p2, int L, int R, int l, int r, int rt){
	if(l>R || r<L) return; 
	p.t1[rt] = max(p1.t1[rt], p2.t1[rt]);
	p.t2[rt] = min(p1.t2[rt], p2.t2[rt]);
	if(l == r) return;
	int m = l + r >> 1;
	merge(p, p1, p2, L, R, l, m, rt<<1);
	merge(p, p1, p2, L, R, m+1, r, rt<<1|1);
}

void build(int l, int r, int rt){
	if(l == r){
		t[rt].build(l, 1, n, 1);
		return;
	}
	int m = l + r >> 1;
	build(l, m, rt<<1);
	build(m+1, r, rt<<1|1);
	merge(t[rt], t[rt<<1], t[rt<<1|1], 1, n, 1, n, 1);
}
void update(int x, int y, int xx, int yy, ll v, int l, int r, int rt){
	t[rt].update(y, yy, v, 1, n, 1);
	if(l==x && r==xx){
		lz[rt].update(y, yy, v, 1, n, 1);
		return;
	}
	int m = l + r >> 1;
	if(m >= xx) update(x, y, xx, yy, v, l, m, rt<<1);
	else if(x > m) update(x, y, xx, yy, v, m+1, r, rt<<1|1);
	else update(x, y, m, yy, v, l, m, rt<<1), \
		 update(m+1, y, xx, yy, v, m+1, r, rt<<1|1);
	merge(t[rt], t[rt<<1], t[rt<<1|1], y, yy, 1, n, 1);
}
ll query1(int x, int y, int xx, int yy, int l, int r, int rt, ll sum){
	if(l==x && r==xx) return t[rt].query1(y, yy, 1, n, 1, 0);
//	sum += lz[rt].query(y, yy, 1, n, 1, 0);
	int m = l + r >> 1;
	if(m >= xx) return query1(x, y, xx, yy, l, m, rt<<1, sum);
	else if(x > m) return query1(x, y, xx, yy, m+1, r, rt<<1|1, sum);
	else return max( query1(x, y, m, yy, l, m, rt<<1, sum), \
					 query1(m+1, y, xx, yy, m+1, r, rt<<1|1, sum) );
}
ll query2(int x, int y, int xx, int yy, int l, int r, int rt, ll sum){
	if(l==x && r==xx) return t[rt].query2(y, yy, 1, n, 1, 0);
//	sum += lz[rt].query(y, yy, 1, n, 1, 0);
	int m = l + r >> 1;
	if(m >= xx) return query2(x, y, xx, yy, l, m, rt<<1, sum);
	else if(x > m) return query2(x, y, xx, yy, m+1, r, rt<<1|1, sum);
	else return min( query2(x, y, m, yy, l, m, rt<<1, sum), \
					 query2(m+1, y, xx, yy, m+1, r, rt<<1|1, sum) );
}

int main() {
	scanf("%d", &n);
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
			scanf("%lld", &a[i][j]);
	build(1, n, 1);
	scanf("%d", &q);
	while(q--){
		char op; int x, y, xx, yy, w;
		scanf(" %c", &op);
		if(op == 'c') {
			scanf("%d %d %d", &x, &y, &w);
			update(x, y, x, y, w, 1, n, 1);
		} else {
			scanf("%d %d %d %d", &x, &y, &xx, &yy);
			printf("%lld ", query1(x, y, xx, yy, 1, n, 1, 0));
			printf("%lld\n", query2(x, y, xx, yy, 1, n, 1, 0));
		}
	}
}

例题二：Census UVA - 11297

查询区间最大值，用最大值 $+$ $w$ 更新区间
注意这里因为标记可以不断取 $m a x$ ，才可以实现区间更新

// https://www.luogu.com.cn/problem/P3437
#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 2;
int d, s, q;
struct seg{
	int t[maxn<<2], lz[maxn<<2];
	void build(int L, int l, int r, int rt){
		if(l == r) {
//			t[rt] = a[L][r];
			return;
		}
		int m = l + r >> 1;
		build(L, l, m, rt<<1);
		build(L, m+1, r, rt<<1|1);
		t[rt] = max(t[rt<<1], t[rt<<1|1]);
	}
	void update(int L, int R, int x, int l, int r, int rt){
		t[rt] = max(t[rt], x);
		if(l==L && r==R){
			lz[rt] = max(lz[rt], x);
			return;
		}
		int m = l + r >> 1;
		if(m >= R) update(L, R, x, l, m, rt<<1);
		else if(L > m) update(L, R, x, m+1, r, rt<<1|1);
		else update(L, m, x, l, m, rt<<1), \
			 update(m+1, R, x, m+1, r, rt<<1|1);
//		t[rt] = max(t[rt<<1], t[rt<<1|1]);
	}
	int query(int L, int R, int l, int r, int rt){
		if(l==L && r==R) return t[rt];
		int m = l + r >> 1, ret = lz[rt];
		if(m >= R) ret = max(ret, query(L, R, l, m, rt<<1));
		else if(L > m) ret = max(ret, query(L, R, m+1, r, rt<<1|1));
		else ret = max(ret, max( query(L, m, l, m, rt<<1), \
						 query(m+1, R, m+1, r, rt<<1|1) ) );
		return ret;
	}
} t[maxn<<2], lz[maxn<<2]; 

void merge(seg &p, seg &p1, seg &p2, int l, int r, int rt){
	p.t[rt] = max(p1.t[rt], p2.t[rt]);
	if(l == r) return;
	int m = l + r >> 1;
	merge(p, p1, p2, l, m, rt<<1);
	merge(p, p1, p2, m+1, r, rt<<1|1);
}

void build(int l, int r, int rt){
	if(l == r){
		t[rt].build(l, 1, s, 1);
		return;
	}
	int m = l + r >> 1;
	build(l, m, rt<<1);
	build(m+1, r, rt<<1|1);
	merge(t[rt], t[rt<<1], t[rt<<1|1], 1, s, 1);
}
void update(int x, int y, int xx, int yy, int v, int l, int r, int rt){
	t[rt].update(y, yy, v, 1, s, 1);
	if(l==x && r==xx){
		lz[rt].update(y, yy, v, 1, s, 1);
		return;
	}
	int m = l + r >> 1;
	if(m >= xx) update(x, y, xx, yy, v, l, m, rt<<1);
	else if(x > m) update(x, y, xx, yy, v, m+1, r, rt<<1|1);
	else update(x, y, m, yy, v, l, m, rt<<1), \
		 update(m+1, y, xx, yy, v, m+1, r, rt<<1|1);
//	merge(t[rt], t[rt<<1], t[rt<<1|1], 1, s, 1);
}
int query(int x, int y, int xx, int yy, int l, int r, int rt){
	if(l==x && r==xx) return t[rt].query(y, yy, 1, s, 1);
	int m = l + r >> 1, ret = lz[rt].query(y, yy, 1, s, 1);
	if(m >= xx) ret = max(ret, query(x, y, xx, yy, l, m, rt<<1));
	else if(x > m) ret = max(ret, query(x, y, xx, yy, m+1, r, rt<<1|1));
	else ret = max(ret, max( query(x, y, m, yy, l, m, rt<<1), \
					 query(m+1, y, xx, yy, m+1, r, rt<<1|1) ) );
	return ret;
}

int main() {
	scanf("%d%d%d", &d, &s, &q);
	while(q--){
		int di, si, w, x, y;
		scanf("%d %d %d %d %d", &di, &si, &w, &x, &y);
		x++, y++;
		w += query(x, y, x+di-1, y+si-1, 1, d, 1);
		update(x, y, x+di-1, y+si-1, w, 1, d, 1);
	}
	printf("%d\n", query(1, 1, d, s, 1, d, 1));
}

例题三：Mosaic - HDU 4819

建树后，查询区间最大最小值，单点更新

// https://vjudge.net/problem/HDU-4819
#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
using pii = pair <int,int>;
const int maxn = 8e2 + 5;
int T, n, q;
int a[maxn][maxn];
struct seg{
	int t1[maxn<<2], t2[maxn<<2], lz[maxn<<2];
	void build(int L, int l, int r, int rt){
		if(l == r) {
			t1[rt] = t2[rt] = a[L][r];
			return;
		}
		int m = l + r >> 1;
		build(L, l, m, rt<<1);
		build(L, m+1, r, rt<<1|1);
		t1[rt] = max(t1[rt<<1], t1[rt<<1|1]);
		t2[rt] = min(t2[rt<<1], t2[rt<<1|1]);
	}
	void update(int L, int R, ll x, int l, int r, int rt){
		t1[rt] = t2[rt] = x;
		if(l==L && r==R){
			lz[rt] = x;
			return;
		}
		int m = l + r >> 1;
		if(m >= R) update(L, R, x, l, m, rt<<1);
		else if(L > m) update(L, R, x, m+1, r, rt<<1|1);
		else update(L, m, x, l, m, rt<<1), \
			 update(m+1, R, x, m+1, r, rt<<1|1);
		t1[rt] = max(t1[rt<<1], t1[rt<<1|1]);
		t2[rt] = min(t2[rt<<1], t2[rt<<1|1]);
	}
	ll query1(int L, int R, int l, int r, int rt, ll sum){
		if(l==L && r==R) return t1[rt];
//		sum += lz[rt];
		int m = l + r >> 1;
		if(m >= R) return query1(L, R, l, m, rt<<1, sum);
		else if(L > m) return query1(L, R, m+1, r, rt<<1|1, sum);
		else return max( query1(L, m, l, m, rt<<1, sum), \
						 query1(m+1, R, m+1, r, rt<<1|1, sum) );
	}
	ll query2(int L, int R, int l, int r, int rt, ll sum){
		if(l==L && r==R) return t2[rt];
//		sum += lz[rt];
		int m = l + r >> 1;
		if(m >= R) return query2(L, R, l, m, rt<<1, sum);
		else if(L > m) return query2(L, R, m+1, r, rt<<1|1, sum);
		else return min( query2(L, m, l, m, rt<<1, sum), \
						 query2(m+1, R, m+1, r, rt<<1|1, sum) );
	}
} t[maxn<<2], lz[maxn<<2]; 

void merge(seg &p, seg &p1, seg &p2, int L, int R, int l, int r, int rt){
	if(l>R || r<L) return; 
	p.t1[rt] = max(p1.t1[rt], p2.t1[rt]);
	p.t2[rt] = min(p1.t2[rt], p2.t2[rt]);
	if(l == r) return;
	int m = l + r >> 1;
	merge(p, p1, p2, L, R, l, m, rt<<1);
	merge(p, p1, p2, L, R, m+1, r, rt<<1|1);
}

void build(int l, int r, int rt){
	if(l == r){
		t[rt].build(l, 1, n, 1);
		return;
	}
	int m = l + r >> 1;
	build(l, m, rt<<1);
	build(m+1, r, rt<<1|1);
	merge(t[rt], t[rt<<1], t[rt<<1|1], 1, n, 1, n, 1);
}
void update(int x, int y, int xx, int yy, ll v, int l, int r, int rt){
	t[rt].update(y, yy, v, 1, n, 1);
	if(l==x && r==xx){
		lz[rt].update(y, yy, v, 1, n, 1);
		return;
	}
	int m = l + r >> 1;
	if(m >= xx) update(x, y, xx, yy, v, l, m, rt<<1);
	else if(x > m) update(x, y, xx, yy, v, m+1, r, rt<<1|1);
	else update(x, y, m, yy, v, l, m, rt<<1), \
		 update(m+1, y, xx, yy, v, m+1, r, rt<<1|1);
	merge(t[rt], t[rt<<1], t[rt<<1|1], y, yy, 1, n, 1);
}
ll query1(int x, int y, int xx, int yy, int l, int r, int rt, ll sum){
	if(l==x && r==xx) return t[rt].query1(y, yy, 1, n, 1, 0);
//	sum += lz[rt].query(y, yy, 1, n, 1, 0);
	int m = l + r >> 1;
	if(m >= xx) return query1(x, y, xx, yy, l, m, rt<<1, sum);
	else if(x > m) return query1(x, y, xx, yy, m+1, r, rt<<1|1, sum);
	else return max( query1(x, y, m, yy, l, m, rt<<1, sum), \
					 query1(m+1, y, xx, yy, m+1, r, rt<<1|1, sum) );
}
ll query2(int x, int y, int xx, int yy, int l, int r, int rt, ll sum){
	if(l==x && r==xx) return t[rt].query2(y, yy, 1, n, 1, 0);
//	sum += lz[rt].query(y, yy, 1, n, 1, 0);
	int m = l + r >> 1;
	if(m >= xx) return query2(x, y, xx, yy, l, m, rt<<1, sum);
	else if(x > m) return query2(x, y, xx, yy, m+1, r, rt<<1|1, sum);
	else return min( query2(x, y, m, yy, l, m, rt<<1, sum), \
					 query2(m+1, y, xx, yy, m+1, r, rt<<1|1, sum) );
}

int main() {
	scanf("%d", &T);
	for(int cas=1; cas<=T; cas++){
		scanf("%d", &n);
		for(int i=1; i<=n; i++)
			for(int j=1; j<=n; j++)
				scanf("%lld", &a[i][j]);
		build(1, n, 1);
		scanf("%d", &q);
		printf("Case #%d:\n", cas);
		while(q--){
			int x, y, L;
			scanf("%d %d %d", &x, &y, &L);
			int x1 = max(x - L / 2, 1);
			int x2 = min(x + L / 2, n);
			int y1 = max(y - L / 2, 1);
			int y2 = min(y + L / 2, n);
			int mx = query1(x1, y1, x2, y2, 1, n, 1, 0);
			int mi = query2(x1, y1, x2, y2, 1, n, 1, 0);
			int w = (mx + mi) / 2;
			printf("%d\n", w);
			update(x, y, x, y, w, 1, n, 1);
		}
	}
}

例题四：SuperBrother打鼹鼠

单点加，查询区间和
注意这份代码也可以满足区间加，查询区间和

#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
using pii = pair <int,int>;
const int maxn = 1024 + 5;
int n;
ll a[maxn][maxn];
struct seg{
	ll t[maxn<<2], lz[maxn<<2];
	void build(int L, int l, int r, int rt){
		if(l == r) {
			t[rt] = a[L][r];
			return;
		}
		int m = l + r >> 1;
		build(L, l, m, rt<<1);
		build(L, m+1, r, rt<<1|1);
		t[rt] = t[rt<<1] + t[rt<<1|1];
	}
	void update(int L, int R, ll x, int l, int r, int rt){
		if(l>R || r<L) return;
		t[rt] += 1ll * x * (R - L + 1);
		if(l==L && r==R){
			lz[rt] += x;
			return;
		}
		int m = l + r >> 1;
		if(m >= R) update(L, R, x, l, m, rt<<1);
		else if(L > m) update(L, R, x, m+1, r, rt<<1|1);
		else update(L, m, x, l, m, rt<<1), \
			 update(m+1, R, x, m+1, r, rt<<1|1);
	}
	ll query(int L, int R, int l, int r, int rt, ll sum){
		if(l>R || r<L) return 0;
		if(l==L && r==R) return t[rt] + sum * (r - l + 1);
		sum += lz[rt];
		int m = l + r >> 1;
		if(m >= R) return query(L, R, l, m, rt<<1, sum);
		else if(L > m) return query(L, R, m+1, r, rt<<1|1, sum);
		else return query(L, m, l, m, rt<<1, sum) +
					query(m+1, R, m+1, r, rt<<1|1, sum);
	}
} t[maxn<<2], lz[maxn<<2]; 

void merge(seg p, seg p1, seg p2, int l, int r, int rt){
	p.t[rt] = p1.t[rt] + p2.t[rt];
	if(l == r) return;
	int m = l + r >> 1;
	merge(p, p1, p2, l, m, rt<<1);
	merge(p, p1, p2, m+1, r, rt<<1|1);
}

void build(int l, int r, int rt){
	if(l == r){
		t[rt].build(l, 1, n, 1);
		return;
	}
	int m = l + r >> 1;
	build(l, m, rt<<1);
	build(m+1, r, rt<<1|1);
	merge(t[rt], t[rt<<1], t[rt<<1|1], 1, n, 1);
}
void update(int x, int y, int xx, int yy, ll v, int l, int r, int rt){
	if(l>xx || r<x) return;
	t[rt].update(y, yy, 1ll * v * (xx - x + 1), 1, n, 1);
	if(l==x && r==xx){
		lz[rt].update(y, yy, v, 1, n, 1);
		return;
	}
	int m = l + r >> 1;
	if(m >= xx) update(x, y, xx, yy, v, l, m, rt<<1);
	else if(x > m) update(x, y, xx, yy, v, m+1, r, rt<<1|1);
	else update(x, y, m, yy, v, l, m, rt<<1), \
		 update(m+1, y, xx, yy, v, m+1, r, rt<<1|1);
}
ll query(int x, int y, int xx, int yy, int l, int r, int rt, ll sum){
	if(l>xx || r<x) return 0;
	if(l==x && r==xx) return t[rt].query(y, yy, 1, n, 1, 0) + sum * (r - l + 1);
	sum += lz[rt].query(y, yy, 1, n, 1, 0);
	int m = l + r >> 1; ll ret = 0;
	if(m >= xx) return query(x, y, xx, yy, l, m, rt<<1, sum);
	else if(x > m) return query(x, y, xx, yy, m+1, r, rt<<1|1, sum);
	else return query(x, y, m, yy, l, m, rt<<1, sum) +
				query(m+1, y, xx, yy, m+1, r, rt<<1|1, sum) ;
}

int main() {
	scanf("%d", &n);
	while(1){
		int m, x1, y1, x2, y2, w;
		scanf("%d", &m);
		if(m == 3) break;
		if(m == 1){
			scanf("%d%d%d", &x1, &y1, &w);
			x1++, y1++, x2++, y2++;
			update(x1, y1, x1, y1, w, 1, n, 1);
		} else {
			scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
			x1++, y1++, x2++, y2++;
			printf("%lld\n", query(x1, y1, x2, y2, 1, n, 1, 0));
		}
	} 
}