| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 1054752 | VMaksimoski008 | Dynamic Diameter (CEOI19_diameter) | C++17 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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;
struct SegTree {
int n;
vector<ll> tree, lazy;
SegTree(int _n) : n(_n), tree(4*_n+50), lazy(4*_n+50) {}
void push(int u, int tl, int tr) {
if(!lazy[u]) return ;
tree[u] += 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, ll v) {
push(u, tl, tr);
if(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] = max(tree[2*u], tree[2*u+1]);
}
ll query(int u, int tl, int tr, int l, int r) {
if(l > tr || tl > r) return 0;
if(l <= tl && tr <= r) return tree[u];
int tm = (tl + tr) / 2;
return max(query(2*u, tl, tm, l, r), query(2*u+1, tm+1, tr, l, r));
}
void update(int l, int r, ll v) { update(1, 0, n-1, l, r, v); }
ll query(int l, int r) { return query(1, 0, n-1, l, r); }
};
ll n, W;
int d[maxn], par[maxn], in[maxn], out[maxn], T[maxn], mx_dep=0, timer=0; ll val[maxn], dp[maxn][3], dist[maxn];
vector<pll> gshit[maxn];
vector<int> graph[maxn];
void dfs1(int u, int p, int c) {
in[u] = timer++; T[u] = c;
mx_dep = max(mx_dep, d[u]);
for(auto &[v, w] : gshit[u]) {
if(v == p) continue;
d[v] = d[u] + 1;
dist[v] = dist[u] + w;
val[v] = w;
par[v] = u;
dfs1(v, u, (u==1?v:c));
}
out[u] = timer-1;
}
void dfs(int u) {
if(u == 0) return ;
dp[u][0] = dp[u][1] = dp[u][2] = 0;
ll mx1=0, mx2=0;
for(int &v : graph[u]) {
if(v == par[u]) continue;
dp[u][0] = max(dp[u][0], dp[v][0] + val[v]);
dp[u][2] = max(dp[u][2], dp[v][2]);
if(dp[v][0] + val[v] >= mx1) {
mx2 = mx1;
mx1 = dp[v][0] + val[v];
} else if(dp[v][0] + val[v] >= mx2) {
mx2 = dp[v][0] + val[v];
}
}
dp[u][1] = mx1 + mx2;
dp[u][2] = max(dp[u][2], max(dp[u][0], dp[u][1]));
dfs(par[u]);
}
void dfs2(int u, int p) {
ll mx1=0, mx2=0;
for(int &v : graph[u]) {
if(v == p) continue;
dfs2(v, u);
dp[u][0] = max(dp[u][0], dp[v][0] + val[v]);
dp[u][2] = max(dp[u][2], dp[v][2]);
if(dp[v][0] + val[v] >= mx1) {
mx2 = mx1;
mx1 = dp[v][0] + val[v];
} else if(dp[v][0] + val[v] >= mx2) {
mx2 = dp[v][0] + val[v];
}
}
dp[u][1] = mx1 + mx2;
dp[u][2] = max(dp[u][2], max(dp[u][0], dp[u][1]));
}
signed main() {
ios_base::sync_with_stdio(false);
cout.tie(0); cin.tie(0);
int q;
cin >> n >> q >> W;
vector<array<ll, 3> > edges;
for(int i=0; i<n-1; i++) {
ll a, b, w;
cin >> a >> b >> w;
edges.push_back({ a, b, w });
gshit[a].push_back({ b, w });
gshit[b].push_back({ a, w });
graph[a].push_back(b);
graph[b].push_back(a);
}
dfs1(1, 1, 0);
for(auto &[a, b, w] : edges) if(d[a] > d[b]) swap(a, b);
if(mx_dep <= 1) {
multiset<ll> ms;
for(auto &[a, b, w] : edges) ms.insert(w);
ll last = 0;
while(q--) {
ll d, e;
cin >> d >> e;
d = (d + last) % (n - 1);
e = (e + last) % W;
ms.erase(ms.find(edges[d][2]));
edges[d][2] = e;
ms.insert(e);
auto it = --ms.end();
ll ans = *it;
if(it != ms.begin()) ans += *(--it);
cout << ans << '\n';
last = ans;
}
return 0;
}
if(mx_dep <= 20 || (n <= 5000 && q <= 5000)) {
ll last = 0;
dfs2(1, 1);
while(q--) {
ll d, e;
cin >> d >> e;
d = (d + last) % (n - 1);
e = (e + last) % W;
val[edges[d][1]] = e;
dfs(edges[d][0]);
cout << dp[1][2] << '\n';
last = dp[1][2];
}
return 0;
}
//mora niz kecot
SegTree tree(n);
multiset<ll> ms;
for(int i=1; i<=n; i++) tree.update(in[i], in[i], dist[i]);
for(int &v : graph[1]) ms.insert(tree.query(in[v], out[v]));
ll last = 0;
while(q--) {
ll d, e;
cin >> d >> e;
d = (d + last) % (n - 1);
e = (e + last) % W;
int x = edges[d][1];
ms.erase(ms.find(in[T[x]], out[T[x]]));
tree.update(in[x], out[x], -edges[d][2]);
edges[d][2] = e;
tree.update(in[x], out[x], edges[d][2]);
ms.insert(tree.query(in[T[x]], out[T[x]]));
auto it = --ms.end();
ll ans = *it;
if(it != ms.begin()) ans += *(--it);
cout << ans << '\n';
last = ans;
}
return 0;
}