제출 #658114

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
658114		jiahng	낙하산 고리들 (IOI12_rings)	C++14	69 / 100	4097 ms	176140 KiB

rings

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
#define ll int
typedef pair<int,int> pi;
typedef vector <int> vi;
typedef vector <pi> vpi;
typedef pair<pi, ll> pii;
typedef set <ll> si;
typedef long double ld;
#define f first
#define s second
#define mp make_pair
#define FOR(i,s,e) for(int i=s;i<=int(e);++i)
#define DEC(i,s,e) for(int i=s;i>=int(e);--i)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define maxn 1000010
#define INF (ll)1e9
#define MOD 1000000007
typedef pair <vi, int> pvi;
typedef pair <int,pi> ipi;
typedef vector <pii> vpii;
int N;
struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
vi st[maxn];
int d[maxn];
int maxD;
vi adj[maxn];
 
vpi all_edges;
int p[6][maxn];
int edges[6][maxn], sz[6][maxn];
vi cycles[6];
int ncycles[6];
vi pos;
int fl(int i, int x){
    if (p[i][x] == x) return x;
    return p[i][x] = fl(i, p[i][x]);
}
 
void add(int i, int A, int B){
    if (sz[fl(i,A)] > sz[fl(i,B)]) swap(A,B);
    
    if (fl(i, A) != fl(i,B)){
        int pA = fl(i,A), pB = fl(i,B);
        if (edges[i][pA] >= sz[i][pA]) ncycles[i]--;
        if (edges[i][pB] >= sz[i][pB]) ncycles[i]--;
        sz[i][pB] += sz[i][pA];
        edges[i][pB] += edges[i][pA];
        //if (cycles[i].find(pA) != cycles[i].end()) cycles[i].erase(pA);
        p[i][pA] = pB;
    }else if (edges[i][fl(i,A)] >= sz[i][fl(i,A)]) ncycles[i]--;
    edges[i][fl(i,B)]++;
    if (edges[i][fl(i,B)] >= sz[i][fl(i,B)]){
        ncycles[i]++; cycles[i].pb(fl(i,B));
    }
}
int bigx;
int gt3 = 0;

void Init(int N_) {
    N = N_;
    maxD = 0;
    FOR(i,0,5) FOR(j,0,N-1) p[i][j] = j,sz[i][j] = 1,edges[i][j]  =0;
    bigx = -1;
}
void Link(int A, int B) {
    if (d[A] == 3) gt3++;
    if (d[B] == 3) gt3++;
    d[A]++; d[B]++;
    st[d[A]].pb(A); st[d[B]].pb(B);
 
    adj[A].pb(B); adj[B].pb(A);
 
    if (pos.empty() && (d[A] == 3 || d[B] == 3)){
        if (d[A] == 3){
            aFOR(i, adj[A]) pos.pb(i);
            pos.pb(A);
        }else if (d[B] == 3){
            pos.pb(B); aFOR(i, adj[B]) pos.pb(i);
        }
        FOR(i,0,pos.size()-1){
            aFOR(j, all_edges) if (j.f != pos[i] && j.s != pos[i]) add(i,j.f,j.s);
        }
    }
    
 
    if (bigx == -1 && (d[A] >= 4 || d[B] >= 4)){
        if (d[A] >= 4) bigx = A;
        else if (d[B] >= 4) bigx = B;
        aFOR(i, all_edges) if (i.f != bigx && i.s != bigx) add(4,i.f,i.s);
    }
    
    all_edges.pb(pi(A,B));
    maxD = max({maxD, d[A], d[B]});
 
    add(5,A,B);
    FOR(i,0,(int)pos.size()-1) if (A != pos[i] && B != pos[i]) add(i,A,B);
    if (bigx != -1 && A != bigx && B != bigx) add(4, A, B);
}
 
int CountCritical() {
    if (N == 1) return 1;
    if (maxD >= 4){
        if (gt3 > 1) return 0;
        if (st[maxD].size() == 1){
            int x = *st[maxD].begin();
            bool no = 0;
            if (!cycles[4].empty()) no = 1;
            aFOR(i, st[3]) if (i != x && find(all(adj[i]), x) == adj[i].end()){
                no = 1; break;
            }
            if (no) return 0;
            return 1;
        }else return 0;
    }else if (maxD <= 1) return N;
    else if (maxD == 3){
        if (st[3].size() > 4) return 0;
       
        int ans = 0;
        FOR(i,0,pos.size()-1){
            bool no = 0;
            aFOR(j, st[3]) if (pos[i] != j && find(all(adj[j]), pos[i]) == adj[j].end()){
                no = 1; break;
            }
            if (no) continue;
            if (cycles[i].empty()){
                //cout << "! " << i << '\n';
                ans++;
            }
        }
        return ans;
    }else{
        if (ncycles[5] == 0) return N;
        else if (ncycles[5] == 1) return sz[5][fl(5,cycles[5][0])];
        else return 0;
    }
}

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