Codeforces - Array and Operations

題目鏈接：Codeforces - Array and Operations

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很顯然的二分圖。只不過建圖麻煩，其實也不是很麻煩，暴力搞就行。

把因子單獨拿出來建圖。

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AC代碼：

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int inf=0x3f3f3f3f;
const int N=2e4+10,M=1e6+10;
int n,m,a[N],cnt,s,t,h[N]; map<int,int> mp[N];
int head[N],nex[M],to[M],w[M],tot=1;
vector<int> v[N];
inline void ade(int a,int b,int c){to[++tot]=b; nex[tot]=head[a]; w[tot]=c; head[a]=tot;}
inline void add(int a,int b,int c){ade(a,b,c);	ade(b,a,0);}
inline int get(int id,int x){if(!mp[id][x]) mp[id][x]=++cnt; return mp[id][x];}
inline int bfs(){
	queue<int> q; q.push(s); memset(h,0,sizeof h); h[s]=1;
	while(q.size()){
		int u=q.front();	q.pop();
		for(int i=head[u];i;i=nex[i]) if(w[i]&&!h[to[i]]) h[to[i]]=h[u]+1,q.push(to[i]);
	}
	return h[t];
}
int dfs(int x,int f){
	if(x==t)	return f;	int fl=0;
	for(int i=head[x];i&&f;i=nex[i]){
		if(w[i]&&h[to[i]]==h[x]+1){
			int mi=dfs(to[i],min(w[i],f));
			w[i]-=mi,w[i^1]+=mi,fl+=mi,f-=mi;
		} 
	}
	if(!fl)	h[x]=-1;
	return fl;
}
inline int dinic(){
	int res=0;
	while(bfs())	res+=dfs(s,inf);
	return res;
}
signed main(){
	cin>>n>>m; t=N-5; cnt=n;
	for(int i=1;i<=n;i++)	cin>>a[i];
	for(int i=1;i<=n;i++){
		int now=a[i];
		if(i&1){
			add(s,i,inf);
			for(int j=2;j*j<=now;j++)	if(now%j==0){
				int tmp=0;
				while(now%j==0)	tmp++,now/=j;
				add(i,get(i,j),tmp),v[i].push_back(j);
			}
			if(now>1)	add(i,get(i,now),1),v[i].push_back(now);
		}else{
			add(i,t,inf);
			for(int j=2;j*j<=now;j++)	if(now%j==0){
				int tmp=0;
				while(now%j==0)	tmp++,now/=j;
				add(get(i,j),i,tmp),v[i].push_back(j);
			}
			if(now>1)	add(get(i,now),i,1),v[i].push_back(now);
		}
	}
	for(int i=1,x,y;i<=m;i++){
		cin>>x>>y;
		if(x%2==0)	swap(x,y);
		for(auto j:v[x])	if(mp[y][j])	add(get(x,j),get(y,j),inf);
	}
	cout<<dinic();
	return 0;
}