题目传送门
https://lydsy.com/JudgeOnline/problem.php?id=4009
题解
考虑怎样的情况就会有一个链覆盖另一个链。
设被覆盖的链为 \(a - b\)，覆盖的链为 \(x - y\)。假设有 \(dfn[a] < dfn[b], dfn[x] < dfn[y]\)
那么如果 \(a\) 是 \(b\) 的祖先，那么令 \(g\) 为 \(a\) 的子树中包含 \(b\) 的点，那么 \(x, y\) 中有一个点在 \(g\) 的子树外面，一个在 \(b\) 子树里面，即 \(x \in [1, dfn[g] - 1], y \in [dfn[b], dfn[b] + siz[b] - 1]\)，\(x \in [dfn[b], dfn[b] + siz[b] - 1], y \in [dfn[g] + siz[g], n]\)。
否则，那么 \(x, y\) 必然一个在 \(a\) 子树中，一个在 \(b\) 子树中。即 \(x \in [dfn[a], dfn[a] + siz[a] - 1]\)，\(y \in [dfn[b], dfn[b] + siz[b] - 1]\)。
那么我们不妨把上面的问题看成求出包含一个点的矩阵的权值的第 \(k\) 小。
这个问题似乎可以用扫描线+动态主席树来做。（就是带修改区间 \(k\) 小值啊）
但是应该整体二分更好写一写，整体二分以后转化为求出一个点被几个矩形覆盖，也是扫描线+树状数组维护。
但是我似乎已经忘了整体二分怎么写了，然后学了很久，写的时候写出了一堆的 bug，调了一整天才调完。
不过这个题目把询问转化为矩形维护的相关操作是一个很好的思想。
时间复杂度 \(O(q\log^2n)\)。
#include<bits/stdc++.h>
#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back
template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}
typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;
template<typename I> inline void read(I &x) {
int f = 0, c;
while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
x = c & 15;
while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
f ? x = -x : 0;
}
const int N = 40000 + 7;
#define lowbit(x) ((x) & -(x))
int n, m, Q, dfc, cnt;
int dep[N], f[N], siz[N], son[N], top[N], dfn[N], pre[N], ans[N];
std::priority_queue<std::pair<int, pii> > q;
struct Matrix {
int x1, x2, y1, y2, v;
inline bool operator < (const Matrix &b) const { return x1 < b.x1; }
} a[N << 1], a1[N << 1], a2[N << 1];
struct Point {
int x, y, k, *ans;
inline bool operator < (const Point &b) const { return x < b.x; }
} b[N], b1[N], b2[N];
struct Edge { int to, ne; } g[N << 1]; int head[N], tot;
inline void addedge(int x, int y) { g[++tot].to = y, g[tot].ne = head[x], head[x] = tot; }
inline void adde(int x, int y) { addedge(x, y), addedge(y, x); }
inline void dfs1(int x, int fa = 0) {
dep[x] = dep[fa] + 1, f[x] = fa, siz[x] = 1;
for fec(i, x, y) if (y != fa) dfs1(y, x), siz[x] += siz[y], siz[y] > siz[son[x]] && (son[x] = y);
}
inline void dfs2(int x, int pa) {
top[x] = pa, dfn[x] = ++dfc, pre[dfc] = x;
if (!son[x]) return; dfs2(son[x], pa);
for fec(i, x, y) if (y != f[x] && y != son[x]) dfs2(y, y);
}
inline int gson(int x, int p) {
int g = 0;
while (top[x] != top[p]) g = top[x], x = f[g];
return x == p ? g : son[p];
}
inline bool intr(int x, int p) { return dfn[x] >= dfn[p] && dfn[x] <= dfn[p] + siz[p] - 1; }
namespace BIT {
int s[N];
inline void qadd(int x, int k) { for (; x <= n; x += lowbit(x)) s[x] += k; }
inline int qsum(int x) {
int ans = 0;
for (; x; x -= lowbit(x)) ans += s[x];
return ans;
}
inline void qadd(int l, int r, int k) { qadd(l, k), qadd(r + 1, -k); }
}
using BIT::qadd;
using BIT::qsum;
inline void solve(int L, int R, int st, int ed, int l, int r) {
if (l > r) return;
if (L == R) {
for (int i = l; i <= r; ++i) *b[i].ans = L;
return;
}
int M = (L + R) >> 1, n1 = 0, n2 = 0, j = l - 1;
for (int i = st; i <= ed; ++i) {
if (a[i].v <= M) {
qadd(a[i].y1, a[i].y2, 1);
q.push(std::make_pair(-a[i].x2, pii(a[i].y1, a[i].y2)));
}
if (i != ed && a[i].x1 == a[i + 1].x1) continue;
while (j < r && (i == ed || b[j + 1].x < a[i + 1].x1)) {
++j;
while (!q.empty() && -q.top().fi < b[j].x) qadd(q.top().se.fi, q.top().se.se, -1), q.pop();
int cnt = qsum(b[j].y);
if (cnt >= b[j].k) b1[++n1] = b[j];
else b2[++n2] = b[j], b2[n2].k -= cnt;
}
}
while (!q.empty()) qadd(q.top().se.fi, q.top().se.se, -1), q.pop();
int m1 = 0, m2 = 0;
for (int i = st; i <= ed; ++i) if (a[i].v <= M) a1[++m1] = a[i]; else a2[++m2] = a[i];
assert(m1 + m2 == ed - st + 1), assert(n1 + n2 == r - l + 1);
std::copy(a1 + 1, a1 + m1 + 1, a + st), std::copy(a2 + 1, a2 + m2 + 1, a + st + m1);
std::copy(b1 + 1, b1 + n1 + 1, b + l), std::copy(b2 + 1, b2 + n2 + 1, b + l + n1);
solve(L, M, st, st + m1 - 1, l, l + n1 - 1), solve(M + 1, R, st + m1, ed, l + n1, r);
}
inline void work() {
std::sort(a + 1, a + cnt + 1);
std::sort(b + 1, b + Q + 1);
solve(0, 1e9, 1, cnt, 1, Q);
for (int i = 1; i <= Q; ++i) printf("%d\n", ans[i]);
}
inline void init() {
read(n), read(m), read(Q);
int x, y;
for (int i = 1; i < n; ++i) read(x), read(y), adde(x, y);
dfs1(1), dfs2(1, 1);
for (int i = 1; i <= m; ++i) {
int x, y, k;
read(x), read(y), read(k);
assert(x != y);
if (dfn[x] > dfn[y]) std::swap(x, y);
if (intr(y, x)) {
x = gson(y, x);
if (dfn[x] > 1) a[++cnt] = (Matrix){ 1, dfn[x] - 1, dfn[y], dfn[y] + siz[y] - 1, k };
if (dfn[x] + siz[x] - 1 < n) a[++cnt] = (Matrix){ dfn[y], dfn[y] + siz[y] - 1, dfn[x] + siz[x], n, k };
} else a[++cnt] = (Matrix){ dfn[x], dfn[x] + siz[x] - 1, dfn[y], dfn[y] + siz[y] - 1, k };
}
for (int i = 1; i <= Q; ++i) {
int x, y, k;
read(x), read(y), read(k);
if (dfn[x] > dfn[y]) std::swap(x, y);
b[i] = (Point){ dfn[x], dfn[y], k, ans + i };
}
}
int main() {
#ifdef hzhkk
freopen("hkk.in", "r", stdin);
#endif
init();
work();
fclose(stdin), fclose(stdout);
return 0;
}