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>
using namespace std;
#define FAST_IO ios_base::sync_with_stdio(0); cin.tie(0)
#define FOR(i, a, b) for(int i = (a); i <= (b); i++)
#define REP(n) FOR(O, 1, (n))
#define pb push_back
#define f first
#define s second
typedef long double ld;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, pii> piii;
typedef vector<int> vi;
typedef vector<pii> vii;
typedef vector<ll> vl;
typedef vector<piii> viii;
const int MAXN = 300100, MAXK = 30;
//const ll MOD = 998244353;
const ll INF = 4e18;
const ld PI = asin(1) * 2;
priority_queue<ll> q[MAXN];
int p[MAXN];
ll c[MAXN];
vi adj[MAXN];
int n, m;
int bg[MAXN];
void print (int i) {
cout << " queue i=" << i << ": ";
priority_queue<ll> tmp;
while (!q[i].empty()) {
ll x = q[i].top();
q[i].pop();
tmp.push(x);
cout << " " << x;
}
while (!tmp.empty()) {
ll x = tmp.top();
tmp.pop();
q[i].push(x);
//cout << " " << x;
}
cout << endl;
}
void dfs (int v) {
// cout << " v = " << v << endl;
bg[v] = v;
if (n+1 <= v) {
REP(2) q[v].push(c[v]);
//print(v);
return;
}
for (int u : adj[v]) {
dfs(u);
if ((int)q[bg[u]].size() > (int)q[bg[v]].size()) bg[v] = bg[u];
/*if ((int)q[u].size() > (int)q[v].size()) swap(q[u], q[v]);
while (!q[u].empty()) {
q[v].push(q[u].top());
q[u].pop();
}*/
}
for (int u : adj[v]) {
//dfs(u);
//if ((int)q[bg[u]].size() > (int)q[bg[v]].size()) bg[v] = bg[u];
//if ((int)q[u].size() > (int)q[v].size()) swap(q[u], q[v]);
if (bg[u] == bg[v]) continue;
while (!q[bg[u]].empty()) {
q[bg[v]].push(q[bg[u]].top());
q[bg[u]].pop();
}
}
int sz = (int)adj[v].size();
REP(sz-1) q[bg[v]].pop();
int fr, to;
to = q[bg[v]].top();
q[bg[v]].pop();
fr = q[bg[v]].top();
q[bg[v]].pop();
q[bg[v]].push(fr+c[v]);
q[bg[v]].push(to+c[v]);
// print(v);
}
int main()
{
FAST_IO;
cin >> n >> m;
FOR(i, 2, n+m) {
cin >> p[i] >> c[i];
adj[p[i]].pb(i);
}
/*FOR(i, 1, n+m) {
cout << " i = " << i << " adj: ";
for (int x : adj[i]) cout << x << " ";
cout << endl;
}*/
dfs (1);
ll cur = 0;
FOR(i, 1, n+m) cur += c[i];
q[bg[1]].pop();
while (!q[bg[1]].empty()) {
cur -= q[bg[1]].top();
q[bg[1]].pop();
}
cout << cur << "\n";
return 0;
}
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