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hackercup / 2023 /round3 /double_stars.cpp
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wjomlex
2023 Problems
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#include <iostream>
#include <map>
#include <vector>
using namespace std;
// Solution:
// - For each edge u-v, find d(u, v) = longest arm length starting with u-v-..
// - For each edge u-v, consider all double stars (x, y) with u-v as the center.
// - Iterate through f[u] and f[v] with 2-pointer, with d' initially 1:
// - For the first (d, x), sum min(degree(u) - 1, degree(v) - 1) for every
// [d', d] (those d's, we can make 2-star with that number of arms).
// - Remove the first and decrease x from degree(u) or degree(v), depending
// on what the pair belongs to.
// - Set d' = d + 1
vector<vector<int>> adj;
vector<vector<int>> d; // d[u][i] = greatest distance reachable from adj[u][i]
vector<map<int, int>> f; // f[u] = map of (d, x), representing d is the
// greatest distance of an arm (u, v) for x edges v
int pre(int u, int par = -1, int l = 0) {
int mai = 0;
for (int i = 0; i < (int)adj[u].size(); ++i) {
int v = adj[u][i];
d[u].push_back(-1);
if (v == par) {
continue;
}
int ret = pre(v, u);
mai = max(mai, ret);
d[u][i] = ret;
}
return 1 + mai;
}
void pre2(int u, int par = -1, int farthest_up = 0) {
vector<int> pref(adj[u].size()), suf(adj[u].size());
for (int i = 0; i < (int)adj[u].size(); ++i) {
pref[i] = max(i ? pref[i - 1] : 0, d[u][i]);
}
for (int i = (int)adj[u].size() - 1; i >= 0; --i) {
suf[i] = max(i + 1 < (int)adj[u].size() ? suf[i + 1] : 0, d[u][i]);
}
for (int i = 0; i < (int)adj[u].size(); ++i) {
int v = adj[u][i];
if (v == par) {
d[u][i] = farthest_up;
continue;
}
// Farthest going down here.
int tmp = 1 + max(i ? pref[i - 1] : 0,
i + 1 < (int)adj[u].size() ? suf[i + 1] : 0);
pre2(v, u, max(1 + farthest_up, tmp));
}
}
long long rec(int u, int par = -1) {
long long ans = 0;
for (int i = 0; i < (int)adj[u].size(); ++i) {
int v = adj[u][i];
if (v == par) {
// Calculate for DST centered at (u, par).
f[u][d[u][i]]--;
auto it1 = f[v].begin();
auto it2 = f[u].begin();
int tot1 = (int)adj[v].size() - 1;
int tot2 = (int)adj[u].size() - 1;
int l = 1;
while (it1 != f[v].end() || it2 != f[u].end()) {
if (it1 != f[v].end() &&
(it2 == f[u].end() || it1->first < it2->first)) {
ans += (long long)(it1->first - l + 1) * min(tot1, tot2);
tot1 -= it1->second;
l = it1->first + 1;
it1++;
} else {
ans += (long long)(it2->first - l + 1) * min(tot1, tot2);
tot2 -= it2->second;
l = it2->first + 1;
it2++;
}
}
// assert(tot1 == 0 && tot2 == 0);
f[u][d[u][i]]++;
continue;
}
f[u][d[u][i]]--;
ans += rec(v, u);
f[u][d[u][i]]++;
}
return ans;
}
long long solve() {
int N;
cin >> N;
adj.assign(N + 1, {});
d.assign(N + 1, {});
f.assign(N + 1, {});
for (int i = 0, p; i < N - 1; i++) {
cin >> p;
adj[i + 2].push_back(p);
adj[p].push_back(i + 2);
}
pre(1);
pre2(1);
for (int u = 1; u <= N; ++u) {
for (auto dd : d[u]) {
f[u][dd]++;
}
}
return rec(1);
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int T;
cin >> T;
for (int t = 1; t <= T; t++) {
cout << "Case #" << t << ": " << solve() << endl;
}
return 0;
}