제출 #684308

#제출 시각아이디문제언어결과실행 시간메모리
684308finn__조이터에서 친구를 만드는건 재밌어 (JOI20_joitter2)C++17
100 / 100
936 ms69128 KiB
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

int cmp[100001];
ll sz[100001], ans = 0;
set<int> child[100001], graph[100001], rgraph[100001];
queue<pair<int, int>> to_merge;

void insert_weak_connection(int A, int B)
{
    graph[A].insert(B);
    rgraph[B].insert(A);
    // If there's A new strong connection between A's and B's components, merge them
    if (graph[B].count(A))
        to_merge.push({A, B});
}

int dsu_size(int A) { return child[A].size() + graph[A].size() + rgraph[A].size(); }

int find(int A) { return (A == cmp[A] ? A : cmp[A] = find(cmp[A])); }

void onion(int A, int B)
{
    if (A == B)
        return;
    // Merge the smaller component into the larger
    if (dsu_size(A) < dsu_size(B))
        swap(A, B);

    // Add new contribution
    ans += sz[B] * child[A].size() + sz[A] * child[B].size();

    // DSU stuff
    cmp[B] = A;
    sz[A] += sz[B];

    // Merge children of B into A
    for (int i : child[B])
    {
        if (child[A].count(i))
            ans -= sz[A];
        else
            child[A].insert(i);
    }
    // Erase the connections
    graph[A].erase(B), rgraph[A].erase(B);
    graph[B].erase(A), rgraph[B].erase(A);
    // Merge the weak connections to other components
    for (int i : graph[B])
    {
        rgraph[i].erase(B);
        insert_weak_connection(A, i);
    }
    for (int i : rgraph[B])
    {
        graph[i].erase(B);
        insert_weak_connection(i, A);
    }
}

int main()
{
    cin.tie(0)->sync_with_stdio(0);
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        cmp[i] = i;
        sz[i] = 1;
        child[i].insert(i);
    }
    while (m--)
    {
        int A, B;
        cin >> A >> B;

        B = find(B);
        // If A isn't in B's component and doesn't already follow someone in B's component...
        if (find(A) != B && !child[B].count(A))
        {
            // We insert A as A follower of B and add sz[find(B)] to the answer
            child[B].insert(A);
            ans += sz[B];

            A = find(A);
            // Add connections between components
            insert_weak_connection(A, B);
            // We may have to merge multiple components for each new event
            while (to_merge.size())
            {
                tie(A, B) = to_merge.front();
                to_merge.pop();
                onion(find(A), find(B));
            }
        }
        cout << ans << '\n';
    }
    return 0;
}
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