제출 #1183317

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1183317		lrnnz	낙하산 고리들 (IOI12_rings)	C++20	0 / 100	445 ms	70776 KiB

rings

#include <bits/stdc++.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
 
#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a).size()
#define pb push_back
#define ll long long
#define ui uint64_t
#define ar array
#define us unordered_set
#define cont(set, element) ((set).find(element) != (set).end())
 
/********* DEBUG *********/
 
template <typename T>
void outvec(const vector<T>& Z){
    for (const T& x : Z)
    cout << x << ' ';
    cout << "\n";
}

void printVariable(const any& var) {
    if (!var.has_value()) {
        cout << "null";
        return;
    }

    if (var.type() == typeid(int)) {
        cout << any_cast<int>(var);
    } else if (var.type() == typeid(double)) {
        cout << any_cast<double>(var);
    } else if (var.type() == typeid(float)) {
        cout << any_cast<float>(var);
    } else if (var.type() == typeid(char)) {
        cout << any_cast<char>(var);
    } else if (var.type() == typeid(bool)) {
        cout << (any_cast<bool>(var) ? "true" : "false");
    } else if (var.type() == typeid(string)) {
        cout << any_cast<string>(var);
    } else if (var.type() == typeid(const char*)) {
        cout << any_cast<const char*>(var);
    } else if (var.type() == typeid(long long)) {
        cout << any_cast<long long>(var);
    } else {
        cout << "[unknown type]";
    }
}

template<typename... Args>
void outval(Args... args) {
    vector<any> variables = {args...};
    
    for (size_t i = 0; i < variables.size(); ++i) {
        printVariable(variables[i]);
        if (i != variables.size() - 1) {
            cout << " ";
        }
    }
    cout << "\n";
}

#define nl "\n"
#define sp << " " <<

/********* DEBUG *********/

const ll MOD = 1000000007;
const ll MOD2 = 998244353;
const ll MOD3 = 1000000000;
const ll needMOD = 987654321; 
const ll mxN = 1000005;
const ll inf = 1e18;

/*                              Thoughts
    Cases where there are no criticals:
    - Two unconnected nodes that don't share any neighbours
    - Two unconnected cycles
    - One cycle AND node outside cycle has 3 or more edges that aren't in cycle

    Cases for criticals
    1. Cycles
    - When there is one cycle, critical can only be inside the cycle.
    - If there is a node with 3 or more neighbours, only nodes with 3 or more 
    neighbours in the cycle are critical     
*/

/*                           Implementation
    DSU keep track of cycles, every Link() union A and B, 
    if there is a cycle only nodes in cycle can be removed,
    if there is two cycles only one crtitical node maximum can be removed
    have to remove all nodes with indegree >= 3, else ans is 0


    Keep track of ind >= 3 cnt, also keep track of their nodes, 
*/




// parent of nd i
vector<ll> pars(mxN), cycleCnt(mxN);

// candidates that need and can be removed
set<int> cands;

vector<vector<ll>> adj(mxN);

// amount of critical nodes, current N, amount of cycles, amount of ind3s to remove, need par to be for cand
ll crits, currN, cycles, needremove, needpar;

// DSU
ll findP(ll x){
    if (pars[x] == x)
    return x;

    return pars[x] = findP(pars[x]);
}

void uni(ll x, ll y){
    x = findP(x);
    y = findP(y);

    pars[x] = y;
}

void Init(int N){
    crits=N;
    needpar=-1;
    cycles=needremove=0;
    currN=N;
    for (int i = 0; i < N; i++){
        cycleCnt[i]=0;
        pars[i]=i;
        adj[i].clear();
    }
}

// bfs to cnt cycles
void bfs(ll a, ll b){
    unordered_map<ll, ll> prev;
    queue<ll> q;
    q.push(a);
    prev[a] = -1;

    ll morethantwoc = 0;

    while (q.size()){
        ll curr = q.front(); q.pop();

        for (auto x : adj[curr]){
            if (x == prev[curr])
            continue;

            if (x == b){
                if (++cycleCnt[x] >= 2){
                    if (++morethantwoc >= 2){
                        crits = 0; return;
                    }
                }

                if (cycleCnt[x] == cycles)
                cands.insert(x);

                while (curr != -1){
                    if (++cycleCnt[curr] >= 2){
                        if (++morethantwoc >= 2){
                            crits = 0; return;
                        }
                    }
                    if (cycleCnt[curr] == cycles)
                    cands.insert(curr);
                    curr=prev[curr];
                }
                return;
            }

            if (prev.find(x) == prev.end()){
                prev[x] = curr;
                q.push(x);
            }
        }
    }
}

void Link(int A, int B){
    if (crits == 0)
    return;

    if (findP(A) == findP(B)){
        cycles++;
        bfs(A,B);
        if (crits == 0)
        return;
    }
    uni(A,B);

    adj[A].pb(B);
    adj[B].pb(A);
    
    if (sz(adj[A]) == 3){
        needremove++;
        cands.insert(A);
        for (auto &x : adj[A])
            cands.insert(x);
    }

    if (sz(adj[B]) == 3){
        needremove++;
        cands.insert(B);
        for (auto &x : adj[B])
            cands.insert(x);
    }

    if (needremove)
        crits = sz(cands);

    set<ll> out;

    for (auto &x : cands){
        ll remove = (sz(adj[x]) >= 3);
        for (auto &y : adj[x]){
            if (sz(adj[y]) == 3)
            remove++;
        }

        if (needremove - remove > 0 || (cycleCnt[x] != cycles))
        out.insert(x);
    }

    for (auto &x : out)
        cands.erase(x);
    

    if (needremove)
        crits = sz(cands);

    // cout << "cands: ";
    // for (auto &x : cands)
    //     cout << x << ' ';
    // cout << "\n";
    
}

int CountCritical(){
    return crits;
}

• Subtask 1: 0 / 20
• Subtask 2: 0 / 17
• Subtask 3: 0 / 18
• Subtask 4: 0 / 14
• Subtask 5: 0 / 31