# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
636381 | ghostwriter | 슈퍼트리 잇기 (IOI20_supertrees) | C++14 | 0 ms | 0 KiB |
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 <vector>
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include <debug.h>
#include "grader.cpp"
#endif
#define st first
#define nd second
#define pb push_back
#define pf push_front
#define _pb pop_back
#define _pf pop_front
#define lb lower_bound
#define ub upper_bound
#define mtp make_tuple
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
typedef long long ll; typedef unsigned long long ull;
typedef double db; typedef long double ldb;
typedef pair<int, int> pi; typedef pair<ll, ll> pll;
typedef vector<int> vi; typedef vector<ll> vll; typedef vector<pi> vpi; typedef vector<pll> vpll;
typedef string str;
template<typename T> T gcd(T a, T b) { return (b == 0? a : gcd(b, a % b)); }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
#define FOR(i, l, r) for (int (i) = (l); (i) <= (r); ++(i))
#define FOS(i, r, l) for (int (i) = (r); (i) >= (l); --(i))
#define EACH(i, x) for (auto &(i) : (x))
#define WHILE while
#define file "TEST"
mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
ll rand(ll l, ll r) { return uniform_int_distribution<ll>(l, r)(rd); }
/*
Tran The Bao
CTL - Da Lat
Cay ngay cay dem nhung deo duoc cong nhan
*/
const int N = 1005;
int P[N][N], adj[N][N], p1[N], s[N], c[N], cnt = 0;
bitset<N> d[N];
vi nodes, cc[N];
int getp(int i) { return i == p1[i]? i : p1[i] = getp(p1[i]); }
bool join(int x, int y) {
x = getp(x);
y = getp(y);
if (x == y) return 0;
if (s[y] > s[x]) swap(x, y);
p1[y] = x;
s[x] += s[y];
return 1;
}
void dfs(int u) {
c[u] = 1;
nodes.pb(u);
FOR(v, 0, n - 1) {
if (!P[u][v] || c[v]) continue;
dfs(v);
}
}
void dfs1(int u) {
c[u] = 1;
cc[cnt].pb(u);
FOR(v, 0, n - 1) {
if (P[u][v] != 1 || c[v]) continue;
dfs1(v);
}
}
void dfs2(int u) {
c[u] = 1;
EACH(v, adj[u]) {
if (c[v]) continue;
dfs2(v);
}
}
int construct(std::vector<std::vector<int>> p) {
int n = p.size();
std::vector<std::vector<int>> ans;
for (int i = 0; i < n; i++) {
std::vector<int> row(n, 0);
ans.push_back(row);
}
FOR(i, 0, n - 1)
FOR(j, 0, n - 1)
P[i][j] = p[i][j];
int sub6 = 0;
FOR(i, 0, n - 1)
FOR(j, 0, n - 1)
if (p[i][j] == 3)
sub6 = 1;
if (sub6) return 0;
bool sub1 = 1;
EACH(i, p)
EACH(j, i)
if (j != 1)
sub1 = 0;
if (sub1) {
FOR(i, 0, n - 2) ans[i][i + 1] = ans[i + 1][i] = 1;
build(ans);
return 1;
}
bool connectivity = 1;
FOR(i, 0, n - 1)
FOR(j, 0, n - 1)
d[i][j] = (p[i][j]? 1 : 0);
FOR(i, 0, n - 1)
FOR(j, 0, n - 1)
if (!d[i][j]) {
bitset<N> cur = d[i] & d[j];
int cnt = cur.count();
if (cnt) connectivity = 0;
}
if (!connectivity) return 0;
bool sub2 = 1;
EACH(i, p)
EACH(j, i)
if (j != 0 && j != 1)
sub2 = 0;
if (sub2) {
FOR(i, 0, n - 1) {
p1[i] = i;
s[i] = 1;
}
FOR(i, 0, n - 1)
FOR(j, 0, n - 1) {
if (!join(i, j)) continue;
ans[i][j] = ans[j][i] = 1;
}
build(ans);
return 1;
}
bool sub3 = 1;
FOR(i, 0, n - 1)
FOR(j, 0, n - 1)
if (i != j && p[i][j] != 0 && p[i][j] != 2)
sub3 = 0;
if (sub3) {
FOR(i, 0, n - 1) {
if (c[i]) continue;
nodes.clear();
dfs(i);
if (sz(nodes) == 1) continue;
if (sz(nodes) == 2) return 0;
FOR(j, 0, sz(nodes) - 2) ans[nodes[j]][nodes[j + 1]] = ans[nodes[j + 1]][nodes[j]] = 1;
ans[nodes[0]][nodes.back()] = ans[nodes.back()][nodes[0]] = 1;
}
build(ans);
return 1;
}
FOR(i, 0, n - 1) {
if (c[i]) continue;
++cnt;
dfs1(i);
FOR(j, 0, sz(cc[cnt]) - 2) ans[cc[cnt][j]][cc[cnt][j + 1]] = ans[cc[cnt][j + 1]][cc[cnt][j]] = 1;
ans[cc[cnt][0]][cc[cnt].back()] = ans[cc[cnt].back()][cc[cnt][0]] = 1;
}
memset(c, 0, sizeof c);
FOR(i, 1, cnt)
FOR(j, 1, cnt) {
if (i == j) continue;
bool ok = 0;
EACH(x, cc[i])
EACH(y, cc[j])
if (p[x][y] == 2)
ok = 1;
if (!ok) continue;
EACH(x, cc[i])
EACH(y, cc[j])
if (p[x][y] != 2)
ok = 0;
if (!ok) return 0;
adj[i][j] = 1;
}
FOR(i, 1, cnt) {
if (c[i]) continue;
nodes.clear();
dfs2(i);
if (sz(nodes) == 1) continue;
if (sz(nodes) == 2) return 0;
FOR(j, 0, sz(nodes) - 2) ans[nodes[j]][nodes[j + 1]] = ans[nodes[j + 1]][nodes[j]] = 1;
}
build(ans);
return 1;
}