CF 510E(Fox And Dinner-把质数拆成奇数和偶数)-CFANZ编程社区

E. Fox And Dinner

time limit per test

memory limit per test

input

output

n foxes there (include Fox Ciel). The i-th fox is ai

They will have dinner around some round tables. You want to distribute foxes such that:

Each fox is sitting at some table.
Each table has at least 3 foxes sitting around it.
The sum of ages of any two adjacent foxes around each table should be a prime number.

k foxes f1, f2, ..., fk are sitting around table in clockwise order, then for 1 ≤ i ≤ k - 1: fi and fi + 1 are adjacent, and f1 and fk

If it is possible to distribute the foxes in the desired manner, find out a way to do that.

Input

n (3 ≤ n ≤ 200): the number of foxes in this party.

n integers ai (2 ≤ ai ≤ 104).

Output

If it is impossible to do this, output "Impossible".

m (

CF 510E(Fox And Dinner-把质数拆成奇数和偶数)_#include

): the number of tables.

m lines, each line should start with an integer k -=– the number of foxes around that table, and then k

If there are several possible arrangements, output any of them.

Sample test(s)

input

43 4 8 9

output

14 1 2 4 3

input

52 2 2 2 2

output

Impossible

input

122 3 4 5 6 7 8 9 10 11 12 13

output

112 1 2 3 6 5 12 9 8 7 10 11 4

input

242 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

output

36 1 2 3 6 5 4 10 7 8 9 12 15 14 13 16 11 10 8 17 18 23 22 19 20 21 24

Note

In example 1, they can sit around one table, their ages are: 3-8-9-4, adjacent sums are: 11, 17, 13 and 7, all those integers are primes.

In example 2, it is not possible: the sum of 2+2 = 4 is not a prime number.

由于事先规定ai>=2，所以a+b>=4

如果a+b是素数,那它是奇数=奇数+偶数

原题的环就变成 ji,ou,ji,ou。。。这样的，所以年龄中，奇数个数=偶数个数

把奇数，偶数换在2边，做2分图，a+b是素数就连边。

原题转变成一个2分图匹配,每个点要匹配2条边

故：建s，t，s向每个偶数连容量为2的边，同理，奇数向t连

建图，网络流

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (400+10)
#define MAXM ((200+100*100)*2+100000)
#define MAXAi (10000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,a[MAXN],a1[MAXN],a2[MAXN],num1[MAXN],num2[MAXN];
class Max_flow  //dinic+当前弧优化     
{      
public:      
    int n,s,t;      
    int q[MAXN];      
    int edge[MAXM],next[MAXM],pre[MAXN],weight[MAXM],size;      
    void addedge(int u,int v,int w)        
    {        
        edge[++size]=v;        
        weight[size]=w;        
        next[size]=pre[u];        
        pre[u]=size;        
    }        
    void addedge2(int u,int v,int w){addedge(u,v,w),addedge(v,u,0);}       
    bool b[MAXN];      
    int d[MAXN];      
    bool SPFA(int s,int t)        
    {        
        For(i,n) d[i]=INF;      
        MEM(b)      
        d[q[1]=s]=0;b[s]=1;        
        int head=1,tail=1;        
        while (head<=tail)        
        {        
            int now=q[head++];        
            Forp(now)        
            {        
                int &v=edge[p];        
                if (weight[p]&&!b[v])        
                {        
                    d[v]=d[now]+1;        
                    b[v]=1,q[++tail]=v;        
                }        
            }            
        }        
        return b[t];        
    }       
    int iter[MAXN];    
    int dfs(int x,int f)    
    {    
        if (x==t) return f;    
        Forpiter(x)    
        {    
            int v=edge[p];    
            if (weight[p]&&d[x]<d[v])    
            {    
                  int nowflow=dfs(v,min(weight[p],f));    
                  if (nowflow)    
                  {    
                    weight[p]-=nowflow;    
                    weight[p^1]+=nowflow;    
                    return nowflow;    
                  }    
            }    
        }    
        return 0;    
    }    
    int max_flow(int s,int t)    
    {    
        int flow=0;    
        while(SPFA(s,t))    
        {    
            For(i,n) iter[i]=pre[i];    
            int f;    
            while (f=dfs(s,INF))    
                flow+=f;     
        }    
        return flow;    
    }     
    void mem(int n,int s,int t)      
    {      
        (*this).n=n;    
        (*this).t=t;      
        (*this).s=s;      
            
        size=1;      
        MEM(pre)     
    }      
}S;      

int p[MAXAi*2]={0},size=0;
bool b[MAXAi*2]={0};
const int N=20000;
void make_prime()
{
    MEM(b)
    Fork(i,2,N)
    {
        if (!b[i]) p[++size]=i;
        For(j,size)
        {
            if (p[j]*i>N) break;  
            b[p[j]*i]=1;
            if (i%p[j]==0) break;
        }
    }    
}
int ans[MAXN][3],st[MAXN][MAXN]={0};
int main()
{
//  freopen("CF510E.in","r",stdin);
//  freopen(".out","w",stdout);
    
    make_prime();
    cin>>n;
    
    int n1=0,n2=0;
    For(i,n)
    {    
        scanf("%d",&a[i]);
        if (a[i]&1) a1[++n1]=a[i],num1[n1]=i;
        else a2[++n2]=a[i],num2[n2]=i; 
    }
    
    if (n1^n2)
    {
        cout<<"Impossible"<<endl;
        return 0;
    }
    else
    {
        const int s=1,t=n+2;
        S.mem(t,s,t);
        For(i,n1) S.addedge2(s,i+1,2);
        For(i,n2) S.addedge2(1+n1+i,t,2);
        
        For(i,n1)
            For(j,n2)
                if (!b[a1[i]+a2[j]])
                    S.addedge2(i+1,1+n1+j,1);
        if (S.max_flow(s,t)^n) 
        {
            cout<<"Impossible"<<endl;
            return 0;
        }
        else
        {
            int tot=0; 
            For(i,n1)
            {
                for(int p=S.pre[i+1];p;p=S.next[p])
                {
                    if (S.weight[p]==0) ans[++tot][1]=num1[i],ans[tot][2]=num2[S.edge[p]-1-n1]; 
                }
            }
            MEM(b)
            
        //  For(i,tot) cout<<ans[i][1]<<' '<<ans[i][2]<<endl;
            
            int m=0;
            For(i,tot)
                if (!b[i])
                {
                    b[i]=1;
                    st[++m][1]=ans[i][1],st[m][2]=ans[i][2];
                    int st_size=2;
                    
                    while (st[m][1]^st[m][st_size])
                    {
                        For(i,tot)
                            if (!b[i])
                            {
                                if (ans[i][1]==st[m][st_size]) {st[m][++st_size]=ans[i][2];b[i]=1; break;}
                                if (ans[i][2]==st[m][st_size]) {st[m][++st_size]=ans[i][1];b[i]=1; break;}
                            } 
                    }
                    st[m][0]=st_size-1;                                         
                }
            cout<<m<<endl;
            For(i,m)
            {
                cout<<st[i][0];
                For(j,st[i][0]) printf(" %d",st[i][j]);
                putchar('\n');
            }        
        }
        
        
        
    } 
    
    return 0;
}