이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
long long ans;
int n, m, tt, cnt_bcc, tam;
int tin[maxn], low[maxn], subtree[2 * maxn];
vector<int> adj[maxn], bcc[2 * maxn];
stack<int> stk;
void dfs(int on, int par = -1) {
tin[on] = low[on] = ++tt;
stk.push(on);
tam++;
for (int to : adj[on]) if (to != par) {
if (tin[to]) {
low[on] = min(low[on], tin[to]);
} else {
dfs(to, on);
low[on] = min(low[on], low[to]);
if (low[to] >= tin[on]) {
cnt_bcc++;
bcc[on].push_back(n + cnt_bcc);
while (bcc[n + cnt_bcc].empty() || bcc[n + cnt_bcc].back() != to) {
bcc[n + cnt_bcc].push_back(stk.top());
stk.pop();
}
}
}
}
}
void dfs2(int on) {
subtree[on] = (on <= n);
for (int to : bcc[on]) {
dfs2(to); // escolhendo o par (x, y) em cada ponto da minha subtree
subtree[on] += subtree[to];
if (on > n) ans -= 1ll * ((int) bcc[on].size()) * subtree[to] * (subtree[to] - 1);
}
// escolhendo o par (x, y) fora da minha subtree
if (on > n) ans -= 1ll * ((int) bcc[on].size()) * (tam - subtree[on]) * (tam - subtree[on] - 1);
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n >> m;
for (int i = 0; i < m; i++) {
int a, b;
cin >> a >> b;
adj[a].push_back(b);
adj[b].push_back(a);
}
for (int i = 1; i <= n; i++) {
if (!tin[i]) {
tam = 0;
dfs(i);
ans += 1ll * tam * (tam - 1) * (tam - 2);
dfs2(i);
}
}
cout << ans << '\n';
return 0;
}
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