This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
//#define int long long
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int mod = 1e9 + 7;
const int LOG = 20;
const int maxn = 1e5 + 5;
const int seg_sz = 5e6+5;
struct SegTree {
int n;
vector<pii> tree;
vector<int> lazy;
void init(int _n) {
n = _n;
tree.resize(4*_n+5);
lazy.resize(4*_n+5);
build(1, 1, n);
}
pii merge(pii a, pii b) {
if(a.first < b.first) return a;
if(a.first > b.first) return b;
return { a.first, a.second + b.second };
}
void build(int u, int tl, int tr) {
if(tl == tr) {
tree[u].second = tr - tl + 1;
} else {
int tm = (tl + tr) / 2;
build(2*u, tl, tm);
build(2*u+1, tm+1, tr);
tree[u] = merge(tree[2*u], tree[2*u+1]);
}
}
void push(int u, int tl, int tr) {
if(!lazy[u]) return ;
tree[u].first += lazy[u];
if(tl != tr) {
lazy[2*u] += lazy[u];
lazy[2*u+1] += lazy[u];
}
lazy[u] = 0;
}
void update(int u, int tl, int tr, int l, int r, int v) {
push(u, tl, tr);
if(tl > tr || l > tr || tl > r) return ;
if(l <= tl && tr <= r) {
lazy[u] += v;
push(u, tl, tr);
return ;
}
int tm = (tl + tr) / 2;
update(2*u, tl, tm, l, r, v);
update(2*u+1, tm+1, tr, l, r, v);
tree[u] = merge(tree[2*u], tree[2*u+1]);
}
void update(int l, int r, int v) { update(1, 1, n, l, r, v); }
pii query() {
push(1, 1, n);
return tree[1];
}
} tree;
int m;
vector<array<int, 3> > graph[maxn];
vector<int> ans(maxn);
void dfs(int u, int p) {
if(tree.query().first == 0) ans[u] = m - tree.query().second;
else ans[u] = m;
for(auto &[v, l, r] : graph[u]) {
if(v == p) continue;
tree.update(l, r, 1);
dfs(v, u);
tree.update(l, r, -1);
}
}
signed main() {
ios_base::sync_with_stdio(false);
cout.tie(0); cin.tie(0);
int n;
cin >> n >> m;
tree.init(m);
for(int i=0; i<n-1; i++) {
int a, b, l, r;
cin >> a >> b >> l >> r;
graph[a].push_back({ b, l, r });
graph[b].push_back({ a, l, r });
}
dfs(1, 1);
for(int i=2; i<=n; i++) cout << ans[i] << '\n';
return 0;
}
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