CodeForces 510E Fox And Dinner

CodeForces 510E

素数打表+最大流+最大流打印路径

注意到a_i (2 ≤ a_i ≤ 10⁴).那么题中的所有素数一定是一个奇数和一个偶数的和。那么所有的桌子都是偶环

将所有的偶数点连接超级源点，奇数点连接超级汇点，容量都为2.将所有的偶数点与奇数点间建容量为1的边，若满流，则可以满足条件

画个图可以知道这样建图是正确的→_→，怎么能想到这一点呢？弱不造

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int maxn=208;
const int maxm=20008;
const int INF=0x7f7f7f7f;
int sb[maxn],sx[maxn];
bool pri[maxm];
void shit()
{
	memset(pri,false,sizeof(pri));
	pri[2]=true;
	int i,j,x;
	for(i=3;i<=20000;i++)
	{
		x=sqrt(i);
		for(j=2;j<=x;j++)
			if(i%j==0)
				break;
		if(j>x)	pri[i]=true;
	}
}
struct fuck{
	int u,v,cap,next;
}edge[maxm];
int tol;
int head[maxn];
void init()
{
	tol=0;
	memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)
{
	edge[tol].u=u;
	edge[tol].v=v;
	edge[tol].next=head[u];
	edge[tol].cap=w;
	head[u]=tol++;
	edge[tol].v=u;
	edge[tol].u=v;
	edge[tol].cap=0;
	edge[tol].next=head[v];
	head[v]=tol++;
}
bool vis[maxn];
int dep[maxn],gap[maxn];
void bfs(int sour,int sink)
{
	queue<int>	q;
	memset(vis,false,sizeof(vis));
	memset(dep,-1,sizeof(dep));
	memset(gap,0,sizeof(gap));
	q.push(sink);
	dep[sink]=0;gap[0]=1;vis[sink]=true;
	int u,v,i;
	while(!q.empty())
	{
		u=q.front();q.pop();
		for(i=head[u];i!=-1;i=edge[i].next)
		{
			v=edge[i].v;
			if(!vis[v]&&edge[i].cap==0)
			{
				vis[v]=true;
				dep[v]=dep[u]+1;
				q.push(v);
				gap[dep[v]]++;
			}
		}
	}
}
int cur[maxn],S[maxm];
int last;
int sap(int sour,int sink)
{
	bfs(sour,sink);
	int max_flow=0,i,u;
	for(i=0;i<=last;i++)	cur[i]=head[i];
	int top=0;
	u=sour;
	while(dep[sour]<last)
	{
		if(u==sink)
		{
			int temp=INF,inser;
			for(i=0;i<top;i++)
			{
				if(edge[S[i]].cap<temp)
				{
					temp=edge[S[i]].cap;
					inser=i;
				}
			}
			for(i=0;i<top;i++)
			{
				edge[S[i]].cap-=temp;
				edge[S[i]^1].cap+=temp;
			}
			max_flow+=temp;
			top=inser;
			u=edge[S[top]].u;
		}
		if(u!=sink&&gap[dep[u]-1]==0)	break;
		for(i=cur[u];i!=-1;i=edge[i].next)
			if(edge[i].cap>0&&dep[u]==dep[edge[i].v]+1)
				break;
		if(i!=-1)
		{
			cur[u]=i;
			S[top++]=i;
			u=edge[i].v;
		}
		else
		{
			int mi=last;
			for(i=head[u];i!=-1;i=edge[i].next)
			{
				if(edge[i].cap==0)	continue;
				if(mi>dep[edge[i].v])
				{
					mi=dep[edge[i].v];
					cur[u]=i;
				}
			}
			--gap[dep[u]];
			dep[u]=mi+1;
			++gap[dep[u]];
			if(u!=sour)	u=edge[S[--top]].u;
		}
	}
	return max_flow;
}
int zhan[maxn];
int lin[maxn];
int sum;
int li[maxn];
int dian[maxn];
int w[maxn];
void dfs(int u,int idx)
{
	vis[u]=true;
	sum++;
	int i;
	if(u<=idx)
	{
		for(i=head[u];i!=-1;i=edge[i].next)
			if(edge[i].cap==0&&edge[i].v!=0&&!vis[edge[i].v])
			{
				lin[u]=edge[i].v;
				dfs(edge[i].v,idx);
			}
	}
	else
	{
		for(i=head[u];i!=-1;i=edge[i].next)
			if(edge[i].cap>0&&edge[i].v!=last-1&&!vis[edge[i].v])
			{
				lin[u]=edge[i].v;
				dfs(edge[i].v,idx);
			}
	}
}
int main()
{
	int i,j,n,m,x;
	shit();
	while(scanf("%d",&n)==1)
	{
		int idx,idb;
		idx=idb=0;
		init();
		for(i=1;i<=n;i++)
		{
			scanf("%d",&x);
			if(x&1)
			{
				sb[++idb]=x;
				dian[idb]=i;
			}
			else
			{
				sx[++idx]=x;
				w[idx]=i;
			}
		}
		for(i=1;i<=idx;i++)
			for(j=1;j<=idb;j++)
			{
				if(pri[sx[i]+sb[j]])
					addedge(i,j+idx,1);
			}
		int sour,sink;
		sour=0;sink=n+1;
		for(i=1;i<=idx;i++)
			addedge(sour,i,2);
		for(i=1;i<=idb;i++)
			addedge(i+idx,sink,2);
		last=n+2;
		int ans=sap(sour,sink);
		if(ans!=n)	printf("Impossible\n");
		else
		{
			ans=0;
			int id=0;
			memset(vis,false,sizeof(vis));
			memset(lin,-1,sizeof(lin));
			for(i=1;i<=idx;i++)
			{
				if(!vis[i])
				{
					sum=0;
					dfs(i,idx);
					li[id]=i;
					zhan[id++]=sum;
				}
			}
			printf("%d\n",id);
			for(i=0;i<id;i++)
			{
				printf("%d",zhan[i]);
				int pa=li[i];
				while(pa!=-1)
				{
					if(pa<=idx)
						printf(" %d",w[pa]);
					else
						printf(" %d",dian[pa-idx]);
					pa=lin[pa];
				}
				printf("\n");
			}
		}
	}
	return 0;
}