Submission #397287

#	Time	Username	Problem	Language	Result	Execution time	Memory
397287		osaaateiasavtnl	Duathlon (APIO18_duathlon)	C++14	23 / 100	180 ms	65340 KiB

count_triplets

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
#include <fstream>
using namespace std;
#define int long long
#define ii pair <int, int>
#define app push_back
#define all(a) a.begin(), a.end()
#define bp __builtin_popcountll
#define ll long long
#define mp make_pair
#define x first
#define y second
#define Time (double)clock()/CLOCKS_PER_SEC
#define debug(x) std::cerr << #x << ": " << x << '\n';
#define FOR(i, n) for (int i = 0; i < n; ++i)
#define FORN(i, n) for (int i = 1; i <= n; ++i)
#define pb push_back
#define trav(a, x) for (auto& a : x)
using vi = vector<int>;
template <typename T>
std::istream& operator >>(std::istream& input, std::vector<T>& data)
{
    for (T& x : data)
        input >> x;
    return input;
}
template <typename T>
std::ostream& operator <<(std::ostream& output, const pair <T, T> & data)
{
    output << "(" << data.x << "," << data.y << ")";
    return output;
}
template <typename T>
std::ostream& operator <<(std::ostream& output, const std::vector<T>& data)
{
    for (const T& x : data)
        output << x << " ";
    return output;
}
ll div_up(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll div_down(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down 
ll math_mod(ll a, ll b) { return a - b * div_down(a, b); }
#define tcT template<class T
#define tcTU tcT, class U
tcT> using V = vector<T>; 
tcT> void re(V<T>& x) { 
    trav(a, x)
        cin >> a;
}
tcT> bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0; 
} // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0; 
}
ll gcd(ll a, ll b) {
    while (b) {
        tie(a, b) = mp(b, a % b);
    }
    return a;
}
const int N = 5e5+7;
int n, m;
vi g[N], tree[N];
int num[N], up[N];
int ptr = 1;
bool cut[N];
void dfs1(int u) {
    up[u] = num[u] = ptr++;
    trav (v, g[u]) {
        if (num[v]) {
            ckmin(up[u], num[v]);
        }
        else {
            tree[u].app(v);
            dfs1(v);
            if (up[v] >= num[u]) {
                cut[v] = 1;
            }
        }
    }
}
int cla;
vi T[N];
void dfs2(int u, int cur) {
    if (cur) {
        T[u].app(cur);
        T[cur].app(u);
    }
    trav (v, tree[u]) {
        if (cut[v]) {
            cla++;
            T[u].app(cla);
            T[cla].app(u);
            dfs2(v, cla);
        }
        else {
            dfs2(v, cur);
        }
    }
}
bool used[N];
int sub[N];
void dfs3(int u, int p) {
    used[u] = 1;
    sub[u] = u <= n;
    trav (v, T[u]) {
        if (v != p) {
            dfs3(v, u);
            sub[u] += sub[v];
        }
    }
}
int ans = 0;
void dfs4(int u, int p, int tot) {
    trav (v, T[u]) {
        if (v != p) {
            dfs4(v, u, tot);
            if (u <= n) {
                int comp = T[v].size();
                ans -= (comp - 1) * (tot - sub[v]) * (tot - sub[v] - 1);
            }
            else {
                int comp = T[u].size();
                ans -= (comp - 1) * sub[v] * (sub[v] - 1);
            }
        }
    }    
}
signed main() {
    #ifdef LOCAL
    #else
    #define endl '\n'
    ios_base::sync_with_stdio(0); cin.tie(0);
    #endif
    cin >> n >> m;
    FOR (i, m) {
        int u, v;
        cin >> u >> v;
        g[u].app(v); g[v].app(u);
    }
    cla = n;
    FORN (u, n) {
        if (!num[u]) {
            dfs1(u);
            dfs2(u, 0);
        }
    }
    FORN (u, n + m) {
        //debug(u);
        //debug(T[u]);
        sort(all(T[u]));
        T[u].resize(unique(all(T[u])) - T[u].begin());
    }
    FORN (u, n) {
        if (!used[u]) {
            dfs3(u, u);
            ans += sub[u] * (sub[u] - 1) * (sub[u] - 2);
            dfs4(u, u, sub[u]);
        }
    }
    cout << ans << endl;
}

• 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