题目来源：Light OJ 1356 Prime Independence

题意：给你n个数 选出最多的数构成一个集合使得任何2个数不是另外一个数的质数倍 x!=k*y 

思路：矛盾的2个数连边 并且所有数分成质因子数为奇数和偶数两部分 以质因子奇偶不同构建二分图 同奇或者同偶的数一定不是另外一个数的质数倍

判断矛盾 首先对每个数因子分解 例如x 有a1个p1质因子 a2个p2质因子...an个pn质因子 x的质因子个数为a1+a2+...+an 记为sum

判断是否存在x/p1  x/p2 ...x/p3 存在并且他们的质因子个数sum奇偶性不同 连边 HK求最大独立集

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 500010;
const int maxm = 40010;
const int INF = 999999999;
struct edge
{
	int v, next;
}e[maxn];
bool vis[maxn];
int prime[maxm];
int a[maxm];
int b[maxn];
int p[maxm];
int num[maxm];
int first[maxm], cnt;

int cx[maxm], cy[maxm];
int dx[maxm], dy[maxm];
int n, dis;
void sieve(int n)
{
	int m = sqrt(n+0.5);
	memset(vis, 0, sizeof(vis));
	vis[0] = vis[1] = 1;
	for(int i = 2; i <= m; i++)
		if(!vis[i])
			for(int j = i*i; j <= n; j += i)
				vis[j] = 1;
}

int get_primes(int n)
{
	sieve(n);
	int c = 0;
	for(int i = 2; i <= n; i++)
		if(!vis[i])
			prime[c++] = i;
	return c;
}
void AddEdge(int u, int v)
{
	e[cnt].v = v;
	e[cnt].next = first[u];
	first[u] = cnt++;
}

bool search()
{
	dis = INF;
	memset(dx, -1, sizeof(dx));
	memset(dy, -1, sizeof(dy));
	queue <int> Q;
	for(int i = 1; i <= n; i++)
	{
		if(cx[i] == -1 && (num[i]&1))
		{
			Q.push(i);
			dx[i] = 0;
		}
	}
	while(!Q.empty())
	{
		int u = Q.front(); Q.pop();
		if(dx[u] > dis)
			break;
		for(int i = first[u]; i != -1; i = e[i].next)
		{
			int v = e[i].v;
			if(dy[v] == -1)
			{
				dy[v] = dx[u] + 1;
				if(cy[v] == -1)
				{
					dis = dy[v];
				}
				else
				{
					dx[cy[v]] = dy[v] + 1;
					Q.push(cy[v]);
				}
			}
		}
	}
	return dis != INF;
}

bool dfs(int u)
{
	for(int i = first[u]; i != -1; i = e[i].next)
	{
		int v = e[i].v;
		if(dy[v] == dx[u] + 1)
		{
			dy[v] = 0;
			if(cy[v] == -1 || dfs(cy[v]))
			{
				cx[u] = v;
				cy[v] = u;
				//printf("**%d %d\n", u, v);
				return true;
			}
		}
	}
	return false;
}
int match()
{
	int ans = 0;
	memset(cx, -1, sizeof(cx));
	memset(cy, -1, sizeof(cy));
	while(search())
	{
		for(int i = 1; i <= n; i++)
		{
			if(cx[i] == -1 && (num[i]&1) && dfs(i))
				ans++;
		}
	}
	return ans;
}
int main()
{
	int cas = 1;
	int T;
	scanf("%d", &T);
	int c = get_primes(500000);
	while(T--)
	{		
		scanf("%d", &n);
		memset(b, 0, sizeof(b));
		memset(num, 0, sizeof(num));
		memset(first, -1, sizeof(first));
		cnt = 0;
		for(int i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
		}
		sort(a+1, a+n+1);
		for(int i = 1; i <= n; i++)
		{
			b[a[i]] = i;
		}
		for(int i = 1; i <= n; i++)
		{
			int x = a[i];
			int sum = 0;
			int sum2 = 0;
			for(int j = 0; j < c && prime[j]*prime[j] <= x; j++)
			{
				if(x % prime[j] == 0)
				{
					p[sum++] = prime[j];
					while(x % prime[j] == 0)
					{
						x /= prime[j];
						sum2++;
					}
				}
			}
			if(x > 1)
			{
				p[sum++] = x;
				sum2++;
			}
			num[i] = sum2;
			//printf("***%d\n", sum2);
			for(int j = 0; j < sum; j++)
			{
				int temp = b[a[i]/p[j]];
				if(temp > i)
					continue;
				if(temp)
				{
					if((sum2&1)==(num[temp]&1))
						continue;			
					//printf("%d %d\n", i, temp);
					if((sum2&1))
						AddEdge(i, temp);
					else
						AddEdge(temp, i);
				}
			}
		}
		int ans = match();
		printf("Case %d: %d\n", cas++, n-ans);
	}
	return 0;
}