Fox And Dinner CodeForces - 510E (最大流)

大意: n只狐狸, 要求分成若干个环, 每个环的狐狸不少于三只, 相邻狐狸年龄和为素数.

狐狸年龄都>=2, 那么素数一定为奇数, 相邻必须是一奇一偶, 也就是一个二分图, 源点向奇数点连容量为2的边, 偶数点向汇点连容量为2的边, 和为偶数的奇数点向偶数点连容量为1的边, 看能否满流即可.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

const int N = 1e5+10;
int n, m, S, T, cnt;
struct _ {
	int from,to,w;
	void pr() {
		printf("fro=%d,to=%d,w=%d\n",from,to,w);
	}
};
vector<_> E;
vector<int> g[N];
int a[N], pre[N], vis[N], d[N];

void add(int x, int y, int w) {
	g[x].pb(E.size());
	E.pb({x,y,w});
	g[y].pb(E.size());
	E.pb({y,x,0});
}

void seive(int n) {
	int mx = sqrt(n+0.5);
	vis[1] = 1;
	REP(i,2,mx) if (!vis[i]) {
		for (int j=i*i; j<=n; j+=i) vis[j]=1;
	}
}
void build() {
	S = n+1, T = n+2;
	REP(i,1,n) {
		if (a[i]&1) add(S,i,2);
		else add(i,T,2);
	}
	REP(i,1,n) if (a[i]&1) {
		REP(j,1,n) if (!vis[a[i]+a[j]]) add(i,j,1);
	}
}

int Maxflow(int s, int t) {
    int ans = 0;
    for (; ; ) {
        REP(i,1,n+2) d[i] = 0;
        queue<int> q;
        q.push(s);
        d[s] = INF;
        while (!q.empty()) {
            int x = q.front();q.pop();
            for (auto &t:g[x]) {
                auto &e = E[t];
                if (!d[e.to]&&e.w>0) {
                    pre[e.to] = t;
                    d[e.to] = min(d[x], e.w);
                    q.push(e.to);
                }
            }
            if (d[t]) break;
        }
        if (!d[t]) break;
        for (int u=t; u!=s; u=E[pre[u]].from) {
            E[pre[u]].w -= d[t];
            E[pre[u]^1].w += d[t];
        }
        ans += d[t];
    }
    return ans;
}

vector<int> gg[N];
vector<int> ans[N];

void dfs(int x) {
	if (vis[x]) return;
	vis[x] = 1;
	ans[cnt].pb(x);
	for (int y:gg[x]) dfs(y);
}

int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d", a+i);
	seive(20000);
	build();
	if (Maxflow(S,T)!=n) return puts("Impossible"),0;
	REP(i,1,n) if (a[i]&1^1) {
		for (auto &&j:g[i]) if (E[j].w==1) {
			gg[E[j].from].pb(E[j].to);
			gg[E[j].to].pb(E[j].from);
		}
	}
	memset(vis,0,sizeof vis);
	REP(i,1,n) if (!vis[i]) ++cnt,dfs(i);
	printf("%d\n", cnt);
	REP(i,1,cnt) {
		printf("%d ",int(ans[i].size()));
		for (auto &&t:ans[i]) printf("%d ",t);hr;
	}
}