제출 #157335

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
157335		dolphingarlic	낙하산 고리들 (IOI12_rings)	C++14	100 / 100	1803 ms	98800 KiB

rings

#include <cassert>
#include <cstdio>
#include <vector>

using namespace std;

int const MAXN = 1000000;

int n;
bool quadruplication = 0;

int numcycles = 0;
int cycle_length;  // If numcycles==1, here we store the length of the only
                   // cycle

int other_endpoint[4][MAXN];  // -1 if the node is not an endpoint, otherwise
                              // the other endpoint
vector<int> neighbours[MAXN];

int destroyed[4];  // The destroyed node of each graph (only if
                   // quadruplication==TRUE)
int degree[4][MAXN];
bool islinear[4];  // Whether each graph is linear or not

void Init(int k) {
    n = k;

    for (int i = 0; i < n; ++i) {
        other_endpoint[0][i] = i;
    }
}

void add_new_edge(int x, int y) {
    // Adds an edge in case of quadruplication

    for (int i = 0; i < 4; ++i) {
        // Operating on graph i

        if (!islinear[i]) continue;
        if (x == destroyed[i] || y == destroyed[i]) continue;

        degree[i][x]++;
        degree[i][y]++;

        assert(degree[i][x] <= 3 && degree[i][y] <= 3);

        if (degree[i][x] == 3 || degree[i][y] == 3) {
            islinear[i] = 0;
            continue;
        }

        if (other_endpoint[i][x] == y) {
            // Cycle!

            islinear[i] = 0;
            continue;
        }

        int a = other_endpoint[i][x];
        int b = other_endpoint[i][y];

        other_endpoint[i][x] = -1;
        other_endpoint[i][y] = -1;
        other_endpoint[i][a] = b;
        other_endpoint[i][b] = a;
    }
}

void quadruplicate(int x) {
    quadruplication = 1;

    destroyed[0] = x;
    destroyed[1] = neighbours[x][0];
    destroyed[2] = neighbours[x][1];
    destroyed[3] = neighbours[x][2];

    for (int i = 0; i < 4; ++i) {
        for (int j = 0; j < n; ++j) {
            other_endpoint[i][j] = j;
            degree[i][j] = 0;
        }
    }

    for (int i = 0; i < 4; ++i) {
        islinear[i] = 1;
    }

    for (int k = 0; k < n; ++k) {
        for (vector<int>::iterator j = neighbours[k].begin();
             j != neighbours[k].end(); ++j) {
            if (k < (*j)) add_new_edge(k, (*j));
        }
    }
}

void Link(int xx, int yy) {
    int x = xx;
    int y = yy;

    if (quadruplication == 0) {
        neighbours[x].push_back(y);
        neighbours[y].push_back(x);

        degree[0][x]++;
        degree[0][y]++;

        // If a node has degree 3, only it or its neighbours can be critical. So
        // we can keep track of each of the 4 graphs obtained by removing one of
        // these 4 nodes.
        if (degree[0][x] == 3) {
            quadruplicate(x);
            return;
        }
        if (degree[0][y] == 3) {
            quadruplicate(y);
            return;
        }

        // If their degree is < 3, then they were necessarily endpoints!

        if (other_endpoint[0][x] != y) {  // A longer path is formed
            int a = other_endpoint[0][x];
            int b = other_endpoint[0][y];

            other_endpoint[0][x] = -1;
            other_endpoint[0][y] = -1;
            other_endpoint[0][a] = b;
            other_endpoint[0][b] = a;
        } else {  // A cycle is formed
            numcycles++;
            if (numcycles == 1) {
                int length = 1;
                int previous_node = x;
                int current_node = neighbours[x][0];

                while (current_node != x) {
                    int possibility = neighbours[current_node][0];
                    if (possibility == previous_node)
                        possibility = neighbours[current_node][1];

                    previous_node = current_node;
                    current_node = possibility;

                    length++;
                }

                cycle_length = length;
            }
        }
    }

    else {
        add_new_edge(x, y);
    }
}

int CountCritical() {
    if (quadruplication == 0) {
        switch (numcycles) {
            case 0:
                return n;

            case 1:
                return cycle_length;

            default:
                return 0;
        }

    } else {
        int answer = 0;
        for (int i = 0; i < 4; ++i) {
            if (islinear[i]) answer++;
        }
        return answer;
    }
}

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