이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "split.h"
#include <bits/stdc++.h>
using namespace std;
#define FAST_IO ios_base::sync_with_stdio(0); cin.tie(nullptr)
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define REP(n) FOR(O, 1, (n))
#define f first
#define s second
#define pb push_back
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vii;
typedef vector<ll> vl;
const int MAXN = 100100;
vi adj[MAXN];
vi seq;
bool ch[MAXN];
int col[3];
int n;
void dfs (int v) {
seq.pb(v);
ch[v] = true;
for (int u : adj[v]) {
if (!ch[u])
dfs(u);
}
}
int sz[MAXN];
bool found = false;
pii paint1, paint2;
void dfsTree (int v,int p) {
if (found) return;
sz[v] ++;
for (int u : adj[v]) {
if (u == p) continue;
dfsTree(u,v);
sz[v] += sz[u];
}
if (found) return;
// cout << " v= " << v << " sz=" << sz[v] << " n - sz = " << n-sz[v] << endl;
FOR(i, 0, 2) FOR(j, 0, 2) {
if (i == j) continue;
if (sz[v] >= col[i] && (n-sz[v]) >= col[j]) {
found = true;
paint1 = {v, i};
paint2 = {p, j};
break;
}
}
}
int painted[MAXN];
void paint (int v, int p, int c) {
if (col[c] > 0) {
painted[v] = c+1;
col[c]--;
}
for (int u : adj[v]) {
if (u == p) continue;
paint(u, v, c);
}
}
vector<int> find_split(int N, int a, int b, int c, vector<int> p, vector<int> q) {
n = N;
vector<int> res(n, 0);
col[0] = a, col[1] = b, col[2] = c;
int m = (int)p.size();
FOR(i, 0, m-1) {
int u = p[i], v = q[i];
adj[u].pb(v);
adj[v].pb(u);
}
if (m == n-1) {
dfsTree (0, -1);
if (!found) return res;
FOR(i, 0, 2) {
if (i == paint1.s || i == paint2.s) continue;
FOR(j, 0, n-1) res[j] = i+1;
}
//cout << " paint1: " << paint1.f << " col:" << paint1.s << endl;
//cout << " paint2: " << paint2.f << " col:" << paint2.s << endl;
paint(paint1.f, paint2.f, paint1.s);
paint(paint2.f, paint1.f, paint2.s);
FOR(i, 0, n-1) if (painted[i] > 0) res[i] = painted[i];
return res;
}
int st = 0;
FOR(i, 0, n-1) if ((int)adj[i].size() == 1) st = i;
dfs (st);
FOR(i, 0, a-1) res[seq[i]] = 1;
FOR(i, a, a+b-1) res[seq[i]] = 2;
FOR(i, a+b, a+b+c-1) res[seq[i]] = 3;
return res;
}
/*
9 10
4 2 3
0 1
0 2
0 3
0 4
0 6
0 8
1 7
3 7
4 5
5 6
ans:
1 1 3 1 2 2 3 1 3
6 5
2 2 2
0 1
0 2
0 3
0 4
0 5
ans:
0 0 0 0 0 0
9 8
4 2 3
0 1
0 2
0 3
0 4
0 8
1 7
4 5
5 6
ans:
1 1 3 1 2 2 3 1 3
*/
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