CodeForces 510E Fox And Dinner

CodeForces 510E

素数打表+最大流+最大流打印路径

注意到ai (2 ≤ ai ≤ 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;
}