이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <vector>
#include <map>
using namespace std;
const int maxn = 1e5 + 5;
int n;
int par[maxn];
int vis[maxn];
vector<int> cycle_starts;
vector<int> adj[maxn];
bool dfs_find_cycle(int nd) {
vis[nd] = 1; // current vertex visited
if (vis[par[nd]] == 0) { // if next vertex unvisited dfs_find_cycle into it
if (dfs_find_cycle(par[nd])) {
vis[nd] = 2;
return true;
}
} else if (vis[par[nd]] == 1) { // if visited next vertex in current iteration
cycle_starts.push_back(nd);
vis[nd] = 2;
return true; // back out
}
vis[nd] = 2; // mark current vertex visited
return false;
}
void find_components() {
for (int i = 0; i < n; i++) {
if (vis[i]) continue;
dfs_find_cycle(i);
}
}
void read_input() {
cin >> n;
map<string, int> m;
int cnt = 0;
for (int i = 0; i < n; i++) {
string a, b; cin >> a >> b;
// if (m.find(a) == m.end()) cout << a << ' ' << cnt << endl;
if (m.find(a) == m.end())
m[a] = cnt++;
// if (m.find(b) == m.end()) cout << b << ' ' << cnt << endl;
if (m.find(b) == m.end())
m[b] = cnt++;
par[m[a]] = m[b];
adj[m[a]].push_back(m[b]);
adj[m[b]].push_back(m[a]);
}
}
int cur_root = -1;
int F[maxn], G[maxn];
int f(int x, int par); int g(int x, int par);
int f(int x, int par) {
F[x] = 1;
for (auto i : adj[x])
if (i != par && i != cur_root)
F[x] += g(i, x);
// cout << "f " << x << ' ' << F[x] << endl;
return F[x];
}
int g(int x, int par) {
G[x] = f(x, par) - 1;
int max_fg = 0;
for (auto i : adj[x])
if (i != par && i != cur_root)
max_fg = max(max_fg, f(i, x) - g(i, x));
G[x] += max_fg;
// cout << "g " << x << ' ' << G[x] << endl;
return G[x];
}
int main() {
// freopen("main.in", "r", stdin);
read_input();
if (n & 1) { cout << -1 << endl; return 0; }
find_components();
fill(F, F + maxn, -1);
fill(G, G + maxn, -1);
int sum_root_g = 0;
for (auto i : cycle_starts) {
// cycle starts @ i & size 1
if (i == par[i]) {
cur_root = i;
sum_root_g += g(i, i);
}
// size 2
else if (i == par[par[i]]) {
// cout << sum_root_g << ' ';
// optimal to take arc
// i, par[i]
for (auto j : adj[i])
if (j != par[i])
sum_root_g += g(j, i);
for (auto j : adj[par[i]])
if (j != i)
sum_root_g += g(j, par[j]);
sum_root_g += 2;
// cout << sum_root_g << endl;
}
else {
// cout << i << endl;
// par[i] --> ... --> i --> par[i]
cur_root = par[i];
// take arc
int with_arc = 1;
for (auto j : adj[i])
if (j != par[i])
with_arc += g(j, i);
for (auto j : adj[par[i]])
if (j != i)
with_arc += g(j, par[i]);
// cout << "with arc " << i << ' ' << par[i] << ' ' << with_arc << endl;
int without_arc = g(par[i], i);
// cout << "without arc " << i << ' ' << par[i] << ' ' << without_arc << endl;
sum_root_g += max(with_arc, without_arc);
}
}
cout << n - sum_root_g << endl;
return 0;
}
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |