Prime Independence
LightOJ - 1356A set of integers is called prime independent if none of its member is a prime multiple of another member. An integer a is said to be a prime multiple of b if,
a = b x k (where k is a prime [1])
So, 6 is a prime multiple of 2, but 8 is not. And for example, {2, 8, 17} is prime independent but {2, 8, 16} or {3, 6} are not.
Now, given a set of distinct positive integers, calculate the largest prime independent subset.
Input
Input starts with an integer T (≤ 20), denoting the number of test cases.
Each case starts with an integer N (1 ≤ N ≤ 40000) denoting the size of the set. Next line contains N integers separated by a single space. Each of these N integers are distinct and between 1 and 500000 inclusive.
OutputFor each case, print the case number and the size of the largest prime independent subset.
Sample Input3
5
2 4 8 16 32
5
2 3 4 6 9
3
1 2 3
Sample OutputCase 1: 3
Case 2: 3
Case 3: 2
Note1. An integer is said to be a prime if it's divisible by exactly two distinct integers. First few prime numbers are 2, 3, 5, 7, 11, 13, ...
2. Dataset is huge, use faster I/O methods.
Restriction: If a % b == 0 && a / b = k where k is a prime number, then a and b cannot exist in a set at the same time.
The meaning of the question: give you n numbers, let you find the largest set that satisfies the above constraints, and output the number of elements.
Idea: Build a bipartite graph according to the parity of the number of prime factors of each number, and construct a good graph. Let's run HK, Hungary didn't dare to write, and even my HK didn't dare to use vector.
Mapping: It is not possible to traverse the mapping with a two-layer for loop, and the O(n*n) mapping complexity cannot be afforded. You can find all prime factors p[] of each number a[i], judge whether a[i] / p[] exists, and then build a map according to the parity. In this way, the worst time complexity of mapping is O(n*10), and then adding HK, the total time complexity
O(sqrt(n) * m + n * 10)。
code：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#define MAXN 40000+10
#define MAXM 1000000+10
#define INF 0x3f3f3f3f
#define debug printf("1\n");
using namespace std;
struct Edge{
int to, next;
};
Edge edge[MAXM];
int head[MAXN], edgenum;
int P[500010], num[500010];//Record whether each number is a prime number record the number of prime factors in each number
int a[MAXN];
int id[MAXN];
int oddnum, evennum;
void getP()
{
memset(P, 0, sizeof(P));
for(int i = 2; i <= 500000; i++)
{
if(P[i]) continue;
for(int j = 2*i; j <= 500000; j+=i)
P[j] = 1;
}
P[1] = 1;
}
//vector<int> p[500000+10];
void getPsum()
{
for(int j = 1; j <= 500000; j++)
{
int cnt = 0;
int n = j;
for(int i = 2; i * i <= n; i++)
{
if(n % i == 0)
{
while(n % i == 0)
{
cnt++;
n /= i;
}
}
}
if(n > 1)
cnt++;
num[j] = cnt;
}
}
void init(){
edge = 0;
memset(head, -1, sizeof(head));
}
void addEdge(int u, int v)
{
Edge E1 = {v, head[u]};
edge[edgenum] = E1;
head[u] = edgenum++;
}
int n;
int force[500010];
int p[30], top;
void getprime(int n)
{
top = 0;
for(int i = 2; i * i <= n; i++)
{
if(n % i == 0)
{
p[top++] = i;
while(n % i == 0)
n /= i;
}
}
if(n > 1)
p[top++] = n;
}
void getMap()
{
scanf("%d", &n);
oddnum = evennum = 0;
memset(vis, 0, sizeof(vis));
for(int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
vis[a[i]] = i;//Mark the element has appeared
if(num[a[i]] & 1)
id[i] = ++oddnum;
else
id[i] = ++evennum;
}
init();
for(int i = 1; i <= n; i++)
{
getprime(a[i]);//Process prime factor
for(int j = 0; j < top; j++)
{
int goal = a[i] / p[j];
int index = vis[goal];
if(index)//exist
{
if(num[a[i]] & 1 && num[a[index]] % 2 == 0)
addEdge(id[i], id[index]);
else if(num[a[i]] % 2 == 0 && num[a[index]] & 1)
addEdge(id[index], id[i]);
}
}
}
}
bool used[MAXN];
int dx[MAXN], dy[MAXN];
int mx[MAXN], my[MAXN];
you are DFS (you are)
{
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(!used[v] && dy[v] == dx[u] + 1)
{
used[v] = true;
if(my[v] == -1 || DFS(my[v]))
{
my[v] = u; mx[u] = v;
return 1;
}
}
}
return 0;
}
int kcase = 1;
void HK()
{
memset(mx, -1, sizeof(mx));
memset(my, -1, sizeof(my));
int years = 0;
while(1)
{
bool flag = false;
memset(dx, 0, sizeof(dx));
memset(dy, 0, sizeof(dy));
queue<int> Q;
for(int i = 1; i <= oddnum; i++)
if(mx[i] == -1)
Q.push(i);
while(!Q.empty())
{
int u = Q.front();
Q.pop();
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(!dy[v])
{
dy[v] = dx[u] + 1;
if(my[v] == -1)
flag = true;
else
{
dx[my[v]] = dx[u] + 1;
Q.push(my[v]);
}
}
}
}
if(!flag) break;
memset(used, false, sizeof(used));
for(int i = 1; i <= oddnum; i++)
if(mx[i] == -1)
ans += DFS(i);
}
printf("Case %d: %d\n", kcase++, n - ans);
}
intmain()
{
getP();
getPsum ();
int t;
scanf("%d", &t);
while(t--)
{
getMap();
HK ();
}
return 0;
}