LibreOJ #526. 「LibreOJ β Round #4」子集

Posted 2020-10-05 ZlycerQan

二次联通门 : LibreOJ #526. 「LibreOJ β Round #4」子集

 

 

 

 

 

/*
    LibreOJ #526. 「LibreOJ β Round #4」子集

    考虑一下，若两个数奇偶性相同
    若同为奇数, 那加1后就是偶数, gcd的乘积就一定不是1
    偶数相同

    那么我们把原数中的偶数分为一个集合，奇数分为一个集合
    把互相之间不符合要求的连边

    那么问题就转化为了二分图求最大独立集
*/
#include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>

const int BUF = 12312312;
char Buf[BUF], *buf = Buf;
#define INF 1e9
inline void read (long long &now)
{
    for (now = 0; !isdigit (*buf); ++ buf);
    for (; isdigit (*buf); now = now * 10 + *buf - ‘0‘, ++ buf);
}

#define Max 600
struct E { E *n, *r; int v, f; };
E *list[Max], poor[Max * 10000], *Ta = poor;
int S, T;

class Flow_Type
{
    private :

        int d[Max];

        bool Bfs (int S, int T)
        {
            memset (d, -1, sizeof d); d[S] = 0;
            std :: queue <int> Queue; int now; E *e;
            for (Queue.push (S); !Queue.empty (); Queue.pop ())
            {
                now = Queue.front ();
                for (e = list[now]; e; e = e->n)
                    if (d[e->v] == -1 && e->f)
                    {
                        d[e->v] = d[now] + 1;
                        if (e->v == T) return true;
                        Queue.push (e->v);
                    }
            }
            return d[T] != -1;
        }
        
        int Flowing (int now, int F)
        {
            if (now == T || !F) return F;
            int res = 0, pos; E *e;
            for (e = list[now]; e; e = e->n)
            {
                if (d[e->v] != d[now] + 1 || !e->f) continue;
                pos = Flowing (e->v, e->f < F ? e->f : F);
                if (pos)
                {
                    F -= pos, res += pos, e->f -= pos, e->r->f += pos;
                    if (!F) return res;
                }
            }
            if (res != F) d[now] = -1;
            return res;
        }

    public :

        inline void In (int u, int v)
        {
            ++ Ta, Ta->v = v, Ta->n = list[u], list[u] = Ta, Ta->f = 1;
            ++ Ta, Ta->v = u, Ta->n = list[v], list[v] = Ta, Ta->f = 0;
            list[u]->r = list[v], list[v]->r = list[u];
        }
        
        int Dinic (int S, int T)
        {
            int Answer = 0;
            for (; Bfs (S, T); Answer += Flowing (S, INF));
            return Answer;
        }
};

Flow_Type F;
long long key[Max];
long long Gcd (long long a, long long b) { return !b ? a : Gcd (b, a % b); }
int Main ()
{
    fread (buf, 1, BUF, stdin);
    long long N; read (N); S = N + 1, T = N + 2;
    register int i, j;
    for (i = 1; i <= N; ++ i)
    {
        read (key[i]);
        if (key[i] & 1) F.In (i, T);
        else F.In (S, i);
    }
    for (i = 1; i <= N; ++ i)
    {
        if ((key[i] & 1) == 0)
            for (j = 1; j <= N; ++ j)
                if (key[j] & 1)
                    if (Gcd (key[i], key[j]) == 1 && Gcd (key[i] + 1, key[j] + 1) == 1)
                        F.In (i, j);
    }

    printf ("%d", N - F.Dinic (S, T));

    return 0;
}

int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}