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 <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
template<typename T>
using Prior = priority_queue<T>;
template<typename T>
using prior = priority_queue<T, vector<T>, greater<T>>;
#define X first
#define Y second
#define ALL(x) (x).begin(), (x).end()
#define eb emplace_back
#define pb push_back
const int maxn = 1000 + 5;
vector<int> par1(maxn), cc1[maxn], par2(maxn), cc2[maxn];
int R1(int x) {return par1[x] ^ x ? par1[x] = R1(par1[x]) : x;}
int R2(int x) {return par2[x] ^ x ? par2[x] = R2(par2[x]) : x;}
void U1(int x, int y) {par1[R1(x)] = R1(y);}
void U2(int x, int y) {par2[R2(x)] = R2(y);}
int construct(vector<vector<int>> p) {
iota(ALL(par1), 0), iota(ALL(par2), 0);
int n = p.size();
vector<vector<int>> ans(n, vector<int>(n, 0));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (p[i][j] == 3) return 0;
if (p[i][j] == 1) U1(i, j);
}
}
for (int i = 0; i < n; ++i) {
cc1[R1(i)].eb(i);
for (int j = 0; j < n; ++j) {
if (!p[i][j] and R1(i) == R1(j)) return 0;
}
}
for (int i = 0; i < n; ++i) {
int sz = cc1[i].size();
for (int j = 0; j < sz-1; ++j) {
ans[cc1[i][j]][cc1[i][j+1]] = 1;
ans[cc1[i][j+1]][cc1[i][j]] = 1;
}
}
build(ans);
return 1;
}
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