제출 #830623

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
830623		Johann	Love Polygon (BOI18_polygon)	C++14	100 / 100	155 ms	11504 KiB

polygon

#include <bits/stdc++.h>
using namespace std;

#define vb vector<bool>
#define vi vector<int>
#define vvi vector<vi>

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin >> n;
    // ungerade fertig
    if (n % 2 == 1) { cout << -1; return 0; }

    map<string, int> namen;
    vi beziehungen(n);
    vi degree(n);      // Grad des Knoten δ+ und δ-
    string sa, sb;
    int a, b;
    int idx = 0;
    for (int j = 0; j < n; j++){
        cin >> sa >> sb;
        if (namen.count(sa) == 0) { namen[sa] = idx; idx++; }
        if (namen.count(sb) == 0) { namen[sb] = idx; idx++; }
        a = namen[sa]; b = namen[sb];
        beziehungen[a] = b;
        degree[b]++;
    }
    // for (map<string, int>::iterator it = namen.begin(); it != namen.end(); ++it) {  cout << it->first << " = " << it->second << endl;  }
    // for (int i = 0; i < n; i++){  cout << i << " --> " << beziehungen[i] << " (" << degree[i] << endl;  }

    // Aussortieren, wer in einer Beziehung ist
    // Knoten mit geringsten Grad finden
    // -> Grad = 1: Erstell Beziehung mit Nachbarn

    int new_edges = 0;
    queue<int> q;
    vb inRelation(n, false);
    for (int i = 0; i < n; ++i) {
        if (beziehungen[i] != i && i == beziehungen[beziehungen[i]]) inRelation[i] = true;
        if (degree[i] == 0) q.push(i);
    }
    while (!q.empty()) {
        int s = q.front(); q.pop();
        if (inRelation[s]) continue;
        int t = beziehungen[s];
        if (inRelation[t]) {
            --degree[t];
            ++degree[s];
            inRelation[s] = true;
            ++new_edges;
        } else {
            int u = beziehungen[t];
            --degree[u];
            ++degree[s];
            ++new_edges;
            inRelation[s] = inRelation[t] = true;
            if (degree[u] == 0) q.push(u);
        }
    }

    // cout << endl << "Herumgespielt" << endl;  for (int i = 0; i < n; i++){  cout << i << " --> " << beziehungen[i] << " (" << degree[i] << endl;  }  cout << endl;


    // -> Grad = 2: Eulerkreise, größer als 2: verbleibende Knoten auf Beziehung überprüfen, dann 0 Kanten, sonst 1 pro Person
    // -> grad >= 3: unmöglich, wegen Schubfach

    int clen, running_neighbor;
    for (int i = 0; i < n; i++){
        // falls in Beziehung: keine Umkopplung
        if (inRelation[i]) continue;
        // andernfalls - find cycle length
        clen = 0;
        running_neighbor = i;
        do {
            inRelation[running_neighbor] = true;
            running_neighbor = beziehungen[running_neighbor];
            ++clen;
        } while(running_neighbor != i);
        new_edges += (clen + 1) / 2;
    }

    cout << new_edges << endl;

    return 0;
}

• Subtask 1: 21 / 21
• Subtask 2: 25 / 25
• Subtask 3: 29 / 29
• Subtask 4: 25 / 25