이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "supertrees.h"
#include <map>
#include <vector>
using namespace std;
#define ll long long
#define ff first
#define ss second
#define ln "\n"
#define pll pair<ll, ll>
struct DSU{
vector<ll> p;
ll n;
DSU(ll N){
n=N;
p.resize(n, -1);
}
ll get(ll x){
return p[x]==-1?x:p[x]=get(p[x]);
}
bool unite(ll a, ll b){
a=get(a);
b=get(b);
if (a==b) return 0;
p[a]=b;
return 1;
}
};
int construct(std::vector<std::vector<int>> p) {
int n = (int)p.size();
DSU tr(n);
vector<vector<int>> A(n);
for (int i=0; i<n; i++){
for (int j=0; j<n; j++){
if (i!=j and p[i][j]==2){
tr.unite(i, j);
}
}
}
for (int i=0; i<n; i++){
for (int j=0; j<n; j++){
if (i!=j and p[i][j]==0 and tr.get(i)==tr.get(j)){
return 0;
}
}
}
map<ll, vector<ll>> cc;
for (ll i=0; i<n; i++){
cc[tr.get(i)].push_back(i);
}
vector<vector<int>> ans(n, vector<int>(n, 0));
for (auto &[comp, mem]:cc){
if (mem.size()<=2){
return 0;
}
for (ll i=0; i<(ll)mem.size(); i++){
ll nxt=(i+1)%(ll)(mem.size()), nxt2=(i+2)%(ll)(mem.size());
ans[mem[i]][mem[nxt]]=ans[mem[i]][mem[nxt2]]=ans[mem[nxt]][mem[i]]=ans[mem[nxt2]][mem[i]]=1;
}
}
build(ans);
return 1;
}
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