CodeForces 510E Fox And Dinner

网络流。

原点到偶数连边，容量为2，

奇数到汇点连边，容量为2，

偶数到与之能凑成素数的奇数连边，容量为1

如果奇数个数不等于偶数个数，输出不可能

如果原点到偶数的边不满流，输出不可能

剩下的情况有解：因为一个偶数点选了两个奇数点，一个奇数点被两个偶数点选择，一定能构造出环。

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn = 500 + 10;
const int INF = 0x7FFFFFFF;
struct Edge
{
    int from, to, cap, flow;
    Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f){}
};
vector<Edge>edges;
vector<int>G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
int n, m, s, t;
int num[maxn];

vector<int>g[maxn];
vector<int> ans[maxn];
bool f[maxn];
int block;

void init()
{
    for (int i = 0; i < maxn; i++) G[i].clear();
    edges.clear();
}
void AddEdge(int from, int to, int cap)
{
    edges.push_back(Edge(from, to, cap, 0));
    edges.push_back(Edge(to, from, 0, 0));
    int w = edges.size();
    G[from].push_back(w - 2);
    G[to].push_back(w - 1);
}
bool BFS()
{
    memset(vis, 0, sizeof(vis));
    queue<int>Q;
    Q.push(s);
    d[s] = 0;
    vis[s] = 1;
    while (!Q.empty())
    {
        int x = Q.front();
        Q.pop();
        for (int i = 0; i<G[x].size(); i++)
        {
            Edge e = edges[G[x][i]];
            if (!vis[e.to] && e.cap>e.flow)
            {
                vis[e.to] = 1;
                d[e.to] = d[x] + 1;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}
int DFS(int x, int a)
{
    if (x == t || a == 0)
        return a;
    int flow = 0, f;
    for (int &i = cur[x]; i<G[x].size(); i++)
    {
        Edge e = edges[G[x][i]];
        if (d[x]+1 == d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0)
        {
            edges[G[x][i]].flow+=f;
            edges[G[x][i] ^ 1].flow-=f;
            flow+=f;
            a-=f;
            if(a==0) break;
        }
    }
    if(!flow) d[x] = -1;
    return flow;
}
int dinic(int s, int t)
{
    int flow = 0;
    while (BFS())
    {
        memset(cur, 0, sizeof(cur));
        flow += DFS(s, INF);
    }
    return flow;
}

bool prime(int x)
{
    for(int i=2;i*i<=x;i++)
        if(x%i==0) return 0;
    return 1;
}

void Find(int now)
{
    f[now]=1;
    ans[block].push_back(now);
    for(int i=0;i<g[now].size();i++)
    {
        if(f[g[now][i]]) continue;
        Find(g[now][i]);
    }
}

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++) scanf("%d",&num[i]);
        init();
        s=0;t=n+1; int e1=0,e2=0;
        int flag=0;
        for(int i=1;i<=n;i++)
        {
            if(num[i]%2==0) { e1++; AddEdge(s,i,2); }
            else { e2++; AddEdge(i,t,2); }
        }
        if(e1!=e2) flag=1;
        else
        {
            for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++)
                    if(num[i]%2==0&&num[j]%2==1&&prime(num[i]+num[j]))
                        AddEdge(i,j,1);
            int Flow=dinic(s,t);
            if(Flow!=2*e1) flag=1;
        }
        if(flag==1) printf("Impossible\n");
        else
        {
            block=0; memset(f,0,sizeof f);
            for(int i=0;i<maxn;i++) { g[i].clear(); ans[i].clear(); }
            for(int i=0;i<edges.size();i=i+2)
            {
                if(edges[i].flow==1)
                {
                    int u=edges[i].from;
                    int v=edges[i].to;
                    g[u].push_back(v);
                    g[v].push_back(u);
                }
            }

            for(int i=1;i<=n;i++)
            {
                if(f[i]) continue;
                Find(i); block++;
            }

            printf("%d\n",block);
            for(int i=0;i<block;i++)
            {
                printf("%d ",ans[i].size());
                for(int j=0;j<ans[i].size();j++)
                    printf("%d ",ans[i][j]);
                printf("\n");
            }
        }
    }
    return 0;
}