KDTree(Bzoj2648: SJY摆棋子)

题面

传送门

KDTree

大概就是一个分割\(k\)维空间的数据结构,二叉树

建立:每层选取一维为关键字,把中间的点拿出来,递归左右,有个\(STL\)函数nth_element可以用一下

维护:维护当前这个点的子树的每一维的最大值和最小值,相当于维护了个高维矩形

查询:直接遍历一棵树是\(O(n)\)的,利用一些独特的性质可以剪枝,因题而异

奇技淫巧:

  1. 把坐标绕原点转\(\alpha\)
  2. 定期重构或者像替罪羊树一样,利用平衡因子判断是否需重构
  3. 每次查询到某一层优先选取答案可能最优的一个儿子先递归
  4. 等等

Sol

\(KDTree\)模板题
利用替罪羊树的思想重构

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

const int maxn(1e6 + 5);
const int inf(2e9);
const double alpha(0.75);

int n, op, rt, q, ans, p[maxn], tot, num, cnt;

struct Point{
	int d[2];


	IL int operator <(RG Point a) const{
		return d[op] < a.d[op];
	}
} a[maxn];

struct KDTree{
	int ch[2], d[2], size;
	int mn[2], mx[2];
} tr[maxn];

IL void Chkmax(RG int &x, RG int y){
	if(y > x) x = y;
}

IL void Chkmin(RG int &x, RG int y){
	if(y < x) x = y;
}

IL int NewNode(){
	return tot ? p[tot--] : ++cnt;
}

IL void Update(RG int x){
	RG int ls = tr[x].ch[0], rs = tr[x].ch[1];
	tr[x].mx[0] = tr[x].mn[0] = tr[x].d[0];
	tr[x].mx[1] = tr[x].mn[1] = tr[x].d[1];
	tr[x].size = tr[ls].size + tr[rs].size + 1;
	if(ls){
		Chkmin(tr[x].mn[0], tr[ls].mn[0]), Chkmin(tr[x].mn[1], tr[ls].mn[1]);
		Chkmax(tr[x].mx[0], tr[ls].mx[0]), Chkmax(tr[x].mx[1], tr[ls].mx[1]);
	}
	if(rs){
		Chkmin(tr[x].mn[0], tr[rs].mn[0]), Chkmin(tr[x].mn[1], tr[rs].mn[1]);
		Chkmax(tr[x].mx[0], tr[rs].mx[0]), Chkmax(tr[x].mx[1], tr[rs].mx[1]);
	}
}

IL int Build(RG int l, RG int r, RG int nop){
	op = nop;
	RG int x = (l + r) >> 1, nw = NewNode();
	nth_element(a + l, a + x, a + r + 1);
	tr[nw].d[0] = a[x].d[0], tr[nw].d[1] = a[x].d[1];
	if(l < x) tr[nw].ch[0] = Build(l, x - 1, nop ^ 1);
	if(x < r) tr[nw].ch[1] = Build(x + 1, r, nop ^ 1);
	Update(nw);
	return nw;
}

IL int Check(RG int x){
	return alpha * tr[x].size < tr[tr[x].ch[0]].size || alpha * tr[x].size < tr[tr[x].ch[1]].size;
}

IL void ReCycle(RG int x){
	if(!x) return;
	ReCycle(tr[x].ch[0]), tr[x].ch[0] = 0;
	p[++tot] = x, a[++num].d[0] = tr[x].d[0], a[num].d[1] = tr[x].d[1];
	ReCycle(tr[x].ch[1]), tr[x].ch[1] = 0;
}

IL int ReBuild(RG int x, RG int nop){
	num = 0, ReCycle(x);
	return Build(1, num, nop);
}

IL void Insert(RG int &x, RG int nop, RG Point np){
	if(!x){
		x = NewNode();
		tr[x].d[0] = np.d[0], tr[x].d[1] = np.d[1];
		tr[x].mx[0] = tr[x].mn[0] = tr[x].d[0];
		tr[x].mx[1] = tr[x].mn[1] = tr[x].d[1];
		return;
	}
	if(np.d[nop] < tr[x].d[nop]) Insert(tr[x].ch[0], nop ^ 1, np);
	else Insert(tr[x].ch[1], nop ^ 1, np);
	Update(x);
	if(Check(x)) x = ReBuild(x, nop);
}

IL int Calc(RG int x, RG Point np){
	RG int d = max(np.d[0] - tr[x].mx[0], 0) + max(tr[x].mn[0] - np.d[0], 0);
	return d + max(tr[x].mn[1] - np.d[1], 0) + max(np.d[1] - tr[x].mx[1], 0);
}

IL void Query(RG int x, RG Point np){
	RG int d = abs(tr[x].d[0] - np.d[0]) + abs(tr[x].d[1] - np.d[1]), d1 = inf, d2 = inf;
	Chkmin(ans, d);
	if(tr[x].ch[0]) d1 = Calc(tr[x].ch[0], np);
	if(tr[x].ch[1]) d2 = Calc(tr[x].ch[1], np);
	RG int c = 0;
	if(d1 > d2) swap(d1, d2), c ^= 1;
	if(d1 < ans) Query(tr[x].ch[c], np);
	if(d2 < ans) Query(tr[x].ch[!c], np);
}

int main(){
	n = Input(), q = Input();
	for(RG int i = 1; i <= n; ++i) a[i].d[0] = Input(), a[i].d[1] = Input();
	rt = Build(1, n, 0);
	for(RG int i = 1; i <= q; ++i){
		RG int t = Input(), x = Input(), y = Input();
		if(t == 1) Insert(rt, 0, (Point){x, y});
		else{
			ans = inf;
			Query(rt, (Point){x, y});
			printf("%d\n", ans);
		}
	}
	return 0;
}
posted @ 2018-05-30 21:50  Cyhlnj  阅读(143)  评论(0编辑  收藏  举报