【LG3242】 [HNOI2015]接水果

题面

洛谷

题解

20pts

对于\(n,P,Q\leq 3000\)，暴力判断每条路径的包含关系然后排序\(kth\)即可，复杂度\(O(PQ\log P)\)

另30pts

原树为一条链。

发现对于每个盘子，也就是区间\(x,y\)，那么对于包含这个区间的水果\(u,v\)，要满足\(u\leq x\leq y\leq v\)。

将水果和盘子放在二维平面上一维排序，一维用数据结构维护即可。

100pts

设对于一个点\(x\)，我们\(dfs\)时第一次访问的时间为\(L_x\)，回溯时时间为\(R_x\)。

那么我们下面讨论一下路径之间的包含关系：

对于路径\(u,v\)(\(dep_u<dep_v\)),包含它的路径\(x,y\)有以下情况：

1.\(lca_{u,v}\neq u\)，显然有\(L_u\leq L_x\leq R_u\),\(L_v\leq L_y\leq R_v\)，

可以看作点\((L_x,L_y)\)包含在矩形\(\{(L_u,L_v),(R_u,R_v)\}\)中。

2.\(lca_{u,v}= u\)，设\(w\)为路径\(u,v\)中\(u\)的儿子，那么显然有一个点在\(v\)的子树内，

另一个点在除了\(w\)子树的其他地方,

写成上面那样的关系，就是点\((L_x,L_y)\)在矩形\(\{(1,L_v),(L_w-1,R_v)\}\cup \{(L_v,R_w+1),(R_v,n)\}\)中。

然后对于这个东西，整体二分+扫描线，看有几个在\((L_x,L_y)\)上的权值在\(mid\leq\)，按照整体二分的套路搞即可。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') w = -1, ch = getchar();
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
    return w * data;
}
const int MAX_N = 4e4 + 5;
struct Graph { int to, next; } e[MAX_N << 1]; int fir[MAX_N], e_cnt;
void clearGraph() { memset(fir, -1, sizeof(fir)); e_cnt = 0; }
void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}; fir[u] = e_cnt++; }
int N, P, Q;
int dep[MAX_N], L[MAX_N], R[MAX_N], tim;
int pa[17][MAX_N];
void dfs(int x, int fa) {
	dep[x] = dep[fa] + 1, L[x] = ++tim;
	for (int i = 0; i < 16; i++) pa[i + 1][x] = pa[i][pa[i][x]];
	for (int i = fir[x]; ~i; i = e[i].next) {
		int v = e[i].to; if (v == fa) continue;
		pa[0][v] = x, dfs(v, x);
	}
	R[x] = tim;
}
int LCA(int u, int v) {
	if (dep[u] < dep[v]) swap(u, v);
	for (int i = 16; i >= 0; i--)
		if ((1 << i) <= dep[u] - dep[v]) u = pa[i][u];
	if (u == v) return u;
	for (int i = 16; i >= 0; i--)
		if (pa[i][u] != pa[i][v]) u = pa[i][u], v = pa[i][v];
	return pa[0][u];
}
int Jump(int x, int num) {
	for (int i = 16; i >= 0; i--) if ((num >> i) & 1) x = pa[i][x];
	return x;
}
int q_cnt, p_cnt;
struct Line { int x, _y, y, op, val; } p[MAX_N << 2], lp[MAX_N << 2], rp[MAX_N << 2];
bool operator < (const Line &l, const Line &r) { return l.x < r.x; }
struct Query { int x, y, k, id; } q[MAX_N], lq[MAX_N], rq[MAX_N];
bool operator < (const Query &l, const Query &r) { return l.x < r.x; }
inline int lb(int x) { return x & -x; }
int ans[MAX_N], c[MAX_N];
void add(int x, int v) { while (x <= N) c[x] += v, x += lb(x); }
int sum(int x) { int res = 0; while (x > 0) res += c[x], x -= lb(x); return res; }
int h[MAX_N], cnt = 0;
void Div(int lval, int rval, int sp, int tp, int sq, int tq) {
	if (sp > tp || sq > tq) return ;
	if (lval == rval) {
		for (int i = sq; i <= tq; i++) ans[q[i].id] = h[lval];
		return ;
	}
	int mid = (lval + rval) >> 1;
	int ql = 0, qr = 0, pl = 0, pr = 0, j = sp;
	for (int i = sq; i <= tq; i++) {
		for ( ; j <= tp && p[j].x <= q[i].x; j++) {
			if (p[j].val > h[mid]) rp[++pr] = p[j];
			else add(p[j]._y, p[j].op), add(p[j].y + 1, -p[j].op), lp[++pl] = p[j];
		}
		int tmp = sum(q[i].y);
		if (q[i].k > tmp) q[i].k -= tmp, rq[++qr] = q[i];
		else lq[++ql] = q[i];
	}
	for ( ; j <= tp; j++)
		if (p[j].val > h[mid]) rp[++pr] = p[j];
		else add(p[j]._y, p[j].op), add(p[j].y + 1, -p[j].op), lp[++pl] = p[j];
	for (int i = 1; i <= pl; i++) add(lp[i]._y, -lp[i].op), add(lp[i].y + 1, lp[i].op);
	for (int i = 1; i <= pl; i++) p[i + sp - 1] = lp[i];
	for (int i = 1; i <= pr; i++) p[sp + pl - 1 + i] = rp[i];
	for (int i = 1; i <= ql; i++) q[i + sq - 1] = lq[i];
	for (int i = 1; i <= qr; i++) q[sq + ql - 1 + i] = rq[i];
	Div(lval, mid, sp, sp + pl - 1, sq, sq + ql - 1);
	Div(mid + 1, rval, sp + pl, tp, sq + ql, tq);
} 

int main () {
#ifndef ONLINE_JUDGE
    freopen("cpp.in", "r", stdin);
#endif
	clearGraph();
	N = gi(), P = gi(), Q = gi();
	for (int i = 1; i < N; i++) {
		int u = gi(), v = gi();
		Add_Edge(u, v), Add_Edge(v, u);
	}
	dfs(1, 0);
	for (int i = 1; i <= P; i++) {
		int u = gi(), v = gi(); h[i] = gi();
		if (L[u] > L[v]) swap(u, v);
		int lca = LCA(u, v);
		if (lca == u) {
			int z = Jump(v, dep[v] - dep[u] - 1);
			p[++p_cnt] = (Line){1, L[v], R[v], 1, h[i]};
			p[++p_cnt] = (Line){L[z], L[v], R[v], -1, h[i]};
			if (R[z] < N) {
				p[++p_cnt] = (Line){L[v], R[z] + 1, N, 1, h[i]};
			    p[++p_cnt] = (Line){R[v] + 1, R[z] + 1, N, -1, h[i]};
			}
		} else {
			p[++p_cnt] = (Line){L[u], L[v], R[v], 1, h[i]};
			p[++p_cnt] = (Line){R[u] + 1, L[v], R[v], -1, h[i]};
		}
	}
	sort(&p[1], &p[p_cnt + 1]);
	sort(&h[1], &h[P + 1]); cnt = unique(&h[1], &h[P + 1]) - h - 1;
	while (Q--) {
		int u = gi(), v = gi(), k = gi();
		if (L[u] > L[v]) swap(u, v);
		q[++q_cnt] = (Query){L[u], L[v], k, q_cnt};
	}
	sort(&q[1], &q[q_cnt + 1]);
	Div(1, cnt, 1, p_cnt, 1, q_cnt);
	for (int i = 1; i <= q_cnt; i++) printf("%d\n", ans[i]);
    return 0;
}