【NOI2019模擬2019.6.20】ichi(kruskal重構樹+KD-tree)

Description:

1<=n<=1e5

題解：

首先在子樹裏就是dfs序的一段區間。

那麼路徑最小值>=d的點呢？

很容易想到把點分樹建出來，然後再上面×××

如果套上這個東西的話就變成了$O(log^3)$，還不說空間有多大。

這個其實就是kruskal重構樹的事，模擬時sb了，沒想到kruskal重構樹可以套到這個上面。

滿足路徑最小值>=d的點同樣在kruskal重構樹的一個子樹裏，也用dfs序搞搞。

那麼就是個三維偏序問題，考慮到這題卡空間，最好寫的顯然是KD-tree。

Code：

#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, B = y; i <= B; i ++)
#define ff(i, x, y) for(int i = x, B = y; i <  B; i ++)
#define fd(i, x, y) for(int i = x, B = y; i >= B; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
#define ul unsigned long long
using namespace std;

const int N = 1e5 + 5;

int n, m, ty, x, y, z, op;
ll a[N], la;

struct edge {
	int fi[N * 2], nt[N * 2], to[N * 2], tot;
	void link(int x, int y) {
	 	nt[++ tot] = fi[x], to[tot] = y, fi[x] = tot;
	}
	int fa[N * 2], p[N * 2], q[N * 2], tp;
	void dg(int x) {
		p[x] = ++ tp;
		for(int i = fi[x]; i; i = nt[i]) if(to[i] != fa[x])
			fa[to[i]] = x, dg(to[i]);
		q[x] = tp;
	}
} e1, e2;

struct bb {
	int x, y, z;
} b[N];

void init() {
	scanf("%d %d %d", &n, &m, &ty);
	fo(i, 1, n) scanf("%lld", &a[i]);
	fo(i, 1, n - 1) {
		scanf("%d %d %d", &x, &y, &z);
		e1.link(x, y); e1.link(y, x);
		b[i].x = x; b[i].y = y; b[i].z = z;
	}
	e1.dg(1);
}


int cmp(bb a, bb b) { return a.z > b.z;}

int f[N * 2], tf;
int F(int x) { return f[x] == x ? x : (f[x] = F(f[x]));}

int fa[18][N * 2], fv[18][N * 2];

void build_kruskal_tree() {
	sort(b + 1, b + n, cmp);
	fo(i, 1, 2 * n) f[i] = i;
	tf = n;
	fo(i, 1, n - 1) {
		int x = F(b[i].x), y = F(b[i].y); z = b[i].z;
		tf ++; f[x] = tf; f[y] = tf;
		fa[0][x] = tf; fa[0][y] = tf;
		fv[0][x] = z; fv[0][y] = z;
		e2.link(tf, x); e2.link(tf, y);
	}
	e2.dg(tf);
	fo(j, 1, 17) fo(i, 1, tf) {
		fa[j][i] = fa[j - 1][fa[j - 1][i]];
		fv[j][i] = fv[j - 1][fa[j - 1][i]];
	}
}

int jump(int x, int y) {
	fd(i, 17, 0) if(fa[i][x] && fv[i][x] >= y)
		x = fa[i][x];
	return x;
}

struct P {
	int a[2], i;
} c[N];
int o, tt, rt, tc[N];
int cc(P a, P b) { return a.a[o] < b.a[o];}
struct tree {
	int l, r;
	int mi[2], mx[2], a[2];
	ll lz, s;
} t[N];
#define i0 t[i].l
#define i1 t[i].r
void cmin(int &x, int y) { x > y ? x = y : 0;}
void cmax(int &x, int y) { x < y ? x = y : 0;}
void merge(tree &a, tree b) {
	fo(j, 0, 1) cmin(a.mi[j], b.mi[j]), cmax(a.mx[j], b.mx[j]);
}
void bt(int &i, int x, int y) {
	i = ++ tt;
	int m = x + y >> 1;
	nth_element(c + x, c + m, c + y + 1, cc);
	fo(j, 0, 1) t[i].mi[j] = t[i].mx[j] = t[i].a[j] = c[m].a[j];
	o = !o;
	if(x < m) bt(i0, x, m - 1), merge(t[i], t[i0]);
	if(m < y) bt(i1, m + 1, y), merge(t[i], t[i1]);
	o = !o;
}
void build_kd_tree() {
	fo(i, 1, n) c[i].a[0] = e1.p[i], c[i].a[1] = e2.p[i], c[i].i = i;
	bt(rt, 1, n);
	fo(i, 1, n) tc[c[i].i] = i;
}
void ad(int i, int x) {
	t[i].lz += x, t[i].s += x;
}
void down(int i) {
	if(t[i].lz) ad(i0, t[i].lz), ad(i1, t[i].lz), t[i].lz = 0;
}
int pl[2], pr[2], pw; ll px;
void add(int i, int x, int y) {
	fo(j, 0, 1) if(t[i].mx[j] < pl[j] || t[i].mi[j] > pr[j]) return;
	int ky = 1; fo(j, 0, 1) if(t[i].mi[j] < pl[j] || t[i].mx[j] > pr[j]) ky = 0;
	if(ky) { ad(i, px); return;}
	ky = 1; fo(j, 0, 1) if(t[i].a[j] < pl[j] || t[i].a[j] > pr[j]) ky = 0;
	if(ky) t[i].s += px;
	int m = x + y >> 1; down(i);
	if(x < m) add(i0, x, m - 1);
	if(m < y) add(i1, m + 1, y);
}
void ft(int i, int x, int y) {
	int m = x + y >> 1; down(i);
	if(m == pw) { px = t[i].s; return;}
	if(pw < m) ft(i0, x, m - 1); else ft(i1, m + 1, y);
}

int main() {
	freopen("ichi.in", "r", stdin);
	freopen("ichi.out", "w", stdout);
	init();
	build_kruskal_tree();
	build_kd_tree();
	fo(i, 1, m) {
		scanf("%d", &op);
		if(op == 2) {
			scanf("%d %d %d", &z, &y, &x);
			if(ty) x = (x + la) % n + 1;
			int nx = jump(x, y);
			pl[0] = e1.p[x]; pr[0] = e1.q[x];
			pl[1] = e2.p[nx]; pr[1] = e2.q[nx];
			px = z;
			add(rt, 1, n);
		} else {
			scanf("%d", &x);
			if(ty) x = (x + la) % n + 1;
			px = 0; pw = tc[x];
			ft(1, 1, n);
			la = a[x] + px;
			pp("%lld\n", la);
		}
	}
}