Submission #553205

#	Time	Username	Problem	Language	Result	Execution time	Memory
553205		tranxuanbach	Duathlon (APIO18_duathlon)	C++17	23 / 100	102 ms	52080 KiB

count_triplets

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define endl '\n'
#define fi first
#define se second
#define For(i, l, r) for (auto i = (l); i < (r); i++)
#define ForE(i, l, r) for (auto i = (l); i <= (r); i++)
#define FordE(i, l, r) for (auto i = (l); i >= (r); i--)
#define Fora(v, a) for (auto v: (a))
#define bend(a) (a).begin(), (a).end()
#define isz(a) ((signed)(a).size())

using ll = long long;
using ld = long double;
using pii = pair <int, int>;
using vi = vector <int>;
using vpii = vector <pii>;
using vvi = vector <vi>;

const int N = 5e5 + 5;

int n, m;
vi adj[N];

int ctrtin = 0, tin[N], low[N];
vi st;

int szcpn;
int cntbcc = 0;
vi adjbcc[N];

void dfs(int u, int p){
    tin[u] = low[u] = ++ctrtin;
    st.emplace_back(u);
    szcpn++;
    Fora(v, adj[u]){
        if (v == p){
            continue;
        }
        if (tin[v]){
            low[u] = min(low[u], tin[v]);
        }
        else{
            dfs(v, u);
            if (low[v] >= tin[u]){
                cntbcc++;
                adjbcc[u].emplace_back(n + cntbcc);
                do{
                    int tv = st.back(); st.pop_back();
                    adjbcc[n + cntbcc].emplace_back(tv);
                } while (adjbcc[n + cntbcc].back() != v);
            }
        }
    }
}

ll ans;

int sz[N];

void dfs_bcc(int u){
    sz[u] = (u <= n);
    Fora(v, adjbcc[u]){
        dfs_bcc(v);
        sz[u] += sz[v];
        if (u > n){
            ans -= (ll)isz(adjbcc[u]) * sz[v] * (sz[v] - 1);
        }
    }
    if (u > n){
        ans -= (ll)isz(adjbcc[u]) * (szcpn - sz[u]) * (szcpn - sz[u] - 1);
    }
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    // freopen("KEK.inp", "r", stdin);
    // freopen("KEK.out", "w", stdout);
    cin >> n >> m;
    ForE(i, 1, m){
        int u, v; cin >> u >> v;
        adj[u].emplace_back(v);
        adj[v].emplace_back(u);
    }

    ForE(u, 1, n){
        if (!tin[u]){
            szcpn = 0;
            dfs(u, 0);
            ans += (ll)szcpn * (szcpn - 1) * (szcpn - 2);
            dfs_bcc(u);
        }
    }
    cout << ans << endl;
}

/*
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INPUT:                                            |
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OUTPUT:                                           |
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*/

• Subtask 1: 0 / 5
• Subtask 2: 0 / 11
• Subtask 3: 0 / 8
• Subtask 4: 10 / 10
• Subtask 5: 13 / 13
• Subtask 6: 0 / 15
• Subtask 7: 0 / 20
• Subtask 8: 0 / 8
• Subtask 9: 0 / 10