이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*
* Author: Nonoze
* Created: Sunday 31/03/2024
*/
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long ll;
#define int long long
#ifdef _IN_LOCAL
#include <bits/DebugTemplate.h>
#else
#define dbg(...) ;
#endif
#define sz(x) (int)(x.size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define endl '\n'
#define endlfl '\n' << flush
#define quit(x) cout << x << endl, return;
const int MOD = 1e9+7;
const ll inf = 1e18;
template<typename T> constexpr bool cmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template<typename T> constexpr bool cmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; }
void solve();
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
int tt=1;
// cin >> tt;
while(tt--) solve();
return 0;
}
struct node
{
int left, right;
int sum, lazy;
node() {
sum=0, lazy=-1, left=-1, right=-1;
}
};
node st[100005*64];
int cnt=1;
int query(int root, int left, int right, int qRight) {
if (right<=qRight) return st[root].sum;
int mid=(left+right)/2;
int s1=0, s2=0;
if (st[root].left!=-1) s1=query(st[root].left, left, mid, qRight);
if (st[root].right!=-1 && mid<qRight) s2=query(st[root].right, mid+1, right, qRight);
return s1+s2;
}
void update(int root, int left, int right, int qPoint, int nValue) {
if (left==right) {
st[root].sum+=nValue*left;
return;
}
int mid=(left+right)/2;
if (mid>=qPoint) {
if (st[root].left==-1) st[root].left = cnt++;
update(st[root].left, left, mid, qPoint, nValue);
st[root].sum=st[st[root].left].sum;
if (st[root].right!=-1) st[root].sum+=st[st[root].right].sum;
}
else {
if (st[root].right==-1) st[root].right = cnt++;
update(st[root].right, mid+1, right, qPoint, nValue);
st[root].sum=st[st[root].right].sum;
if (st[root].left!=-1) st[root].sum+=st[st[root].left].sum;
}
}
int n, m, q;
vector<set<pair<int, int>>> adj;
vector<int> depth, nb;
vector<vector<int>> up;
int base=0;
void dfs(int s, int prec=-1) {
for (auto u: adj[s]) {
if (u.fi==prec) continue;
nb[u.fi]=nb[s]+(u.se?1:0);
if (u.se) base=u.se;
depth[u.fi]=depth[s]+1;
up[u.fi][0]=s;
for (int i=1; i<20; i++) {
up[u.fi][i]=up[ up[u.fi][i-1] ][i-1];
}
dfs(u.fi, s);
}
}
int get_lca(int a, int b) {
if (depth[a]<depth[b]) swap(a, b);
int k=depth[a]-depth[b];
for (int i=0; i<20; i++) {
if (k&(1<<i)) {
a=up[a][i];
}
}
if (a==b) return a;
for (int i=19; i>=0; i--) {
if (up[a][i]!=up[b][i]) {
a=up[a][i];
b=up[b][i];
}
}
return up[a][0];
}
void subtask2() {
depth.resize(n), nb.resize(n), up.resize(n, vector<int>(20));
dfs(0);
while (q--) {
int s, t, a, b; cin >> s >> t >> a >> b;
int lca=get_lca(--s, --t);
int sumnb=nb[s]+nb[t]-2*nb[lca];
int take=b/base;
int res=-1;
take=sumnb-take;
if (take<=0) {
cout << a << endl;
continue;
}
cmax(res, a-take);
cout << res << endl;
}
}
int ask(ordered_multiset<int> &ost, int s, int t, int a, int b) {
if (ost.empty()) return a;
int l=0, r=*ost.rbegin(), ans=0;
while (l<=r) {
int mid=(l+r)/2;
if (query(0, 1, (int)1e9, mid)<=b) {
l=mid+1;
ans=mid;
} else r=mid-1;
}
auto ub=ost.lower_bound(ans);
if (ub==ost.end()) return a;
int comp=ost.order_of_key(*ub);
b-=query(0, 1, (int)1e9, ans);
comp+=b/(*ub);
comp=sz(ost)-comp;
if (comp<=0) comp=0;
if (comp>a) comp=a+1;
return a-comp;
}
void subtask3() {
vector<int> ordre(1, 0);
vector<pair<int, int>> nxt(n, {-1, -1});
vector<int> empl(n, 0);
int lst=-1;
for (int i=0; i<n; i++) {
for (auto u: adj[ordre.back()]) {
if (u.fi!=lst) {
nxt[ordre.back()]={u.fi, u.se};
lst=ordre.back();
ordre.push_back(u.fi);
empl[ordre.back()]=sz(ordre)-1;
break;
}
}
}
vector<int> answers(q);
vector<pair<vector<int>, int>> queries;
for (int i=0; i<q; i++) {
int s, t, a, b; cin >> s >> t >> a >> b;
if (--s>--t) swap(s, t);
queries.push_back({{s, t, a, b}, i});
}
sort(all(queries), [&](auto &a, auto &b){
if ((a.fi[0]/300)!=(b.fi[0]/300)) return (a.fi[0]/300)<(b.fi[0]/300);
return a.fi[1]<b.fi[1];
});
int left=empl[queries[0].fi[0]], right=left;
ordered_multiset<int> ost;
for (int i=0; i<q; i++) {
int s=queries[i].fi[0], t=queries[i].fi[1];
while (right<empl[t]) {
if (nxt[ordre[right]].se) {
update(0, 1, (int)1e9, nxt[ordre[right]].se, 1);
ost.insert(nxt[ordre[right]].se);
}
right++;
}
while (left>empl[s]) {
left--;
if (nxt[ordre[left]].se) {
update(0, 1, (int)1e9, nxt[ordre[left]].se, 1);
ost.insert(nxt[ordre[left]].se);
}
}
while (left<empl[s]) {
if (nxt[ordre[left]].se) {
update(0, 1, (int)1e9, nxt[ordre[left]].se, -1);
ost.erase(--ost.lower_bound(nxt[ordre[left]].se));
}
left++;
}
while (right>empl[t]) {
right--;
if (nxt[ordre[right]].se) {
update(0, 1, (int)1e9, nxt[ordre[right]].se, -1);
ost.erase(--ost.lower_bound(nxt[ordre[right]].se));
}
}
answers[queries[i].se]=ask(ost, s, t, queries[i].fi[2], queries[i].fi[3]);
}
for (auto u: answers) cout << u << endl;
}
void solve() {
cin >> n >> m >> q;
adj.resize(n);
bool sub3=true;
vector<pair<int, int>> edges;
for (int i=1; i<n; i++) {
int u, v; cin >> u >> v;
edges.push_back({--u, --v});
if (v!=u+1) sub3=false;
adj[u].insert({v, 0});
adj[v].insert({u, 0});
}
vector<stack<int>> updates(n-1);
for (int i=0; i<m; i++) {
int p, c; cin >> p >> c;
updates[--p].push(c);
}
for (int i=0; i<sz(updates); i++) {
if (updates[i].empty()) continue;
auto act=updates[i];
int u=edges[i].fi, v=edges[i].se;
adj[u].erase({v, 0}); adj[v].erase({u, 0});
adj[u].insert({v, act.top()});
adj[v].insert({u, act.top()});
int weightbase=act.top(); act.pop();
while (!act.empty()) {
int weight=act.top(); act.pop();
adj[u].erase({v, weightbase}); adj[v].erase({u, weightbase});
adj[u].insert({n, weight});
adj.push_back({});
adj[n].insert({u, weight});
adj[v].insert({n, weightbase});
adj[n].insert({v, weightbase});
u=n;
n++;
}
}
if (sub3) return subtask3();
return subtask2();
}
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