题面
Sol
考虑每个盘子怎样才能接到一个水果
分两种情况：
盘子的(x, y)在一条链上，那么水果的两点就要在这条链之外
不在的话，水果的两点就分别在盘子的两点的子树中
记录下每个点的(dfs)序，和这棵子树结束的(dfs)序
发现上述讨论就是相当于求水果这个点((x, y))包含它的二维矩阵(盘子)有多少个
扫描线+树链剖分+树状数组+整体二分
矩阵的样子看代码
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e5 + 5);
IL ll Input(){
RG ll 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;
}
int size[_], fa[_], deep[_], top[_], son[_], dfn[_], Index, ed[_];
int fst[_], nxt[_], to[_], cnt, n;
IL void Add(RG int u, RG int v){
nxt[cnt] = fst[u]; to[cnt] = v; fst[u] = cnt++;
}
IL void Dfs1(RG int u){
size[u] = 1;
for(RG int e = fst[u]; e != -1; e = nxt[e]){
if(size[to[e]]) continue;
fa[to[e]] = u; deep[to[e]] = deep[u] + 1;
Dfs1(to[e]);
size[u] += size[to[e]];
if(size[to[e]] > size[son[u]]) son[u] = to[e];
}
}
IL void Dfs2(RG int u, RG int Top){
dfn[u] = ++Index; top[u] = Top;
if(son[u]) Dfs2(son[u], Top);
for(RG int e = fst[u]; e != -1; e = nxt[e])
if(!dfn[to[e]]) Dfs2(to[e], to[e]);
ed[u] = Index;
}
IL int LCA(RG int u, RG int v, RG int &qs){
while(top[u] ^ top[v]){
if(deep[top[u]] > deep[top[v]]) swap(u, v);
qs = top[v]; v = fa[top[v]];
}
if(deep[u] > deep[v]) swap(u, v);
if(u != v) qs = son[u];
return u;
}
int c[_];
IL void Modify(RG int x, RG int d){
for(; x <= n; x += x & -x) c[x] += d;
}
IL int Query(RG int x){
RG int ret = 0;
for(; x; x -= x & -x) ret += c[x];
return ret;
}
int m, ans[_], Q, tot;
struct Matrix{
int x1, x2, y1, y2, k;
IL bool operator <(RG Matrix B) const{
return k < B.k;
}
} p[_];
struct Option{
int x, l, r, d;
IL bool operator <(RG Option B) const{
return x < B.x;
}
} opt[_];
struct Data{
int id, x, y, k;
IL bool operator <(RG Data B) const{
return x < B.x;
}
} q[_], q1[_], q2[_];
IL void Solve(RG int l, RG int r, RG int L, RG int R){
if(L > R) return;
if(l == r){
for(RG int i = L; i <= R; ++i) ans[q[i].id] = p[l].k;
return;
}
RG int mid = (l + r) >> 1, t0 = 0, t1 = 0, t2 = 0;
for(RG int i = l; i <= mid; ++i){
opt[++t0] = (Option){p[i].x1, p[i].y1, p[i].y2, 1};
opt[++t0] = (Option){p[i].x2 + 1, p[i].y1, p[i].y2, -1};
}
sort(opt + 1, opt + t0 + 1); RG int tt = 1;
for(RG int i = L; i <= R; ++i){
for(; tt <= t0 && opt[tt].x <= q[i].x; ++tt)
Modify(opt[tt].l, opt[tt].d), Modify(opt[tt].r + 1, -opt[tt].d);
RG int s = Query(q[i].y);
if(q[i].k <= s) q1[++t1] = q[i];
else q[i].k -= s, q2[++t2] = q[i];
}
for(; tt <= t0; ++tt) Modify(opt[tt].l, opt[tt].d), Modify(opt[tt].r + 1, -opt[tt].d);
for(RG int i = L, j = 1; j <= t1; ++i, ++j) q[i] = q1[j];
for(RG int i = L + t1, j = 1; j <= t2; ++i, ++j) q[i] = q2[j];
Solve(l, mid, L, L + t1 - 1); Solve(mid + 1, r, L + t1, R);
}
int main(RG int argc, RG char* argv[]){
n = Input(); m = Input(); Q = Input(); Fill(fst, -1);
for(RG int i = 1; i < n; ++i){
RG int u = Input(), v = Input();
Add(u, v); Add(v, u);
}
Dfs1(1); Dfs2(1, 1);
for(RG int i = 1; i <= m; ++i){
RG int x = Input(), y = Input(), k = Input();
if(dfn[x] > dfn[y]) swap(x, y);
RG int qs, lca = LCA(x, y, qs);
if(x == lca){
if(ed[qs] < n) p[++tot] = (Matrix){dfn[y], ed[y], ed[qs] + 1, n, k};
if(dfn[qs] > 1) p[++tot] = (Matrix){1, dfn[qs] - 1, dfn[y], ed[y], k};
}
else p[++tot] = (Matrix){dfn[x], ed[x], dfn[y], ed[y], k};
}
sort(p + 1, p + tot + 1);
for(RG int i = 1; i <= Q; ++i){
RG int x = Input(), y = Input(), k = Input();
if(dfn[x] > dfn[y]) swap(x, y);
q[i] = (Data){i, dfn[x], dfn[y], k};
}
sort(q + 1, q + Q + 1);
Solve(1, tot, 1, Q);
for(RG int i = 1; i <= Q; ++i) printf("%d
", ans[i]);
return 0;
}