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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;
//Dijkstra->set
//set.find_by_order(x) x-position value
//set.order_of_key(x) number of strictly less elements don't need *set.??
#define N 100005
#define wr cout << "Continue debugging\n";
#define all(x) (x).begin(), (x).end()
#define ll long long int
#define pii pair <int, int>
#define pb push_back
#define ff first
#define ss second
vector<int> E[N];
void dfs(int nd, vector<ll>&vis, vector<ll>&sub, ll&sum){
vis[nd] = sub[nd] = 1;
set<int> s;
for (auto i : E[nd]){
if (!vis[i]){
vis[i] = 1;
s.insert(i);
}
}
for (auto i : s){
dfs(i, vis, sub, sum);
sub[nd] += sub[i];
}
sum += sub[nd];
}
int main ()
{
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
for (int i = 1; i <= n; i++){
int k;
cin >> k;
while(k--){
int x;
cin >> x;
E[x].pb(i);
}
}
ll mn = 1e18;
for (int r = 1; r <= n; r++){
ll sum = 0;
vector<ll> vis(n+1, 0), sub(n+1, 0);
dfs(r, vis, sub, sum);
bool tr = 0;
for (int j = 1; j <= n; j++) if (!vis[j]){tr=1;break;}
if (!tr) mn = min(mn, sum);
}
cout << mn << '\n';
}
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