Codeforces Round #290 (Div. 2)E. Fox And Dinner——最大流 奇偶建图

http://codeforces.com/contest/510/problem/E

n只狐狸要吃饭，要求每张桌子至少3只狐狸，相邻两只狐狸的年龄和是素数

因为年龄至少为2，所以和为素数一定是奇数，奇数=奇数+偶数
得到一个二部图
超级源点0对所有的奇数连一条边，cap=2，所有的偶数和超级汇点连一条边，cap=2.如果某个奇数和某个偶数和为素数，则连一条边，cap=1.
满流有解

#include<bits/stdc++.h>
const int maxn=210;
const int maxm=45000;
const int INF=1<<30;
using namespace std;
int num[maxn],d[maxn],p[maxn],cur[maxm];
int s,t;
int n,a[maxn];
int vis[maxn],ans[maxn],cnt;
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];
vector<int> mp[maxn];
inline void addedge(int u,int v,int c)
{
    edges.push_back(Edge(u,v,c,0));
    edges.push_back(Edge(v,u,0,0));
    int m=edges.size();
    G[u].push_back(m-2);
    G[v].push_back(m-1);
}
int Augment(){
    int x=t,a=INF;
    while(x!=s){
        Edge& e=edges[p[x]];
        a=min(a,e.cap-e.flow);
        x=e.from;
    }
    x=t;
    while(x!=s){
        edges[p[x]].flow+=a;
        edges[p[x]^1].flow-=a;
        x=edges[p[x]].from;
    }
    return a;
}
int ISAP()
{
    int flow=0;
    memset(num,0,sizeof(num));
    for(int i=0;i<n+2;++i) num[d[i]]++;
    int x=s;
    memset(cur,0,sizeof(cur));
    while(d[s]<n+2){
        if(x==t){
            flow+=Augment();
            x=s;
        }
        int ok=0;
        for(int i=cur[x];i<G[x].size();++i){
            Edge& e=edges[G[x][i]];
            if(e.cap>e.flow&&d[e.to]+1==d[x]){
                p[e.to]=G[x][i];
                cur[x]=i;
                x=e.to;
                ok=1;
                break;
            }
        }
        if(!ok){
            int m=INF;
            for(int i=0;i<G[x].size();++i){
                Edge& e=edges[G[x][i]];
                if(e.cap>e.flow) m=min(m,d[e.to]);
            }
            if(--num[d[x]]==0) break;
            num[d[x]=m+1]++;
            cur[x]=0;
            if(x!=s) x=edges[p[x]].from;
        }
    }
    return flow;
}
bool isprime(int x)
{
    for(int i=2;i<=sqrt(x);++i){
        if(x%i==0) return false;
    }
    return true;
}
void dfs(int u){
    ans[++cnt]=u;
    for(int i=0;i<mp[u].size();++i){
        int v=mp[u][i];
        if(!vis[v]){
            vis[v]=1;
            dfs(v);
        }
    }
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.cpp","r",stdin);
#endif // ONLINE_JUDGE
    cin>>n;
    for(int i=1;i<=n;++i) scanf("%d",a+i);
    for(int i=1;i<=n;++i){
        if(a[i]&1) addedge(0,i,2);
        else addedge(i,n+1,2);
    }
    for(int i=1;i<=n;++i){
        if(~a[i]&1) continue;
        for(int j=1;j<=n;++j){
            if(a[j]&1) continue;
            if(isprime(a[i]+a[j])) addedge(i,j,1);
        }
    }
    s=0,t=n+1;
    int res=ISAP();
    if(res!=n) printf("Impossible\n");
    else{
        for(int i=1;i<=n;++i){
            if(~a[i]&1) continue;
            for(int j=0;j<G[i].size();++j){
                Edge& e=edges[G[i][j]];
                int v=e.to;
                if(v>0&&v<n+1&&(e.cap==e.flow)){
                    mp[i].push_back(v);
                    mp[v].push_back(i);
                }
            }
        }
        int count=0;
        for(int i=1;i<=n;++i){
            if(!vis[i]){
                ++count;
                vis[i]=1;
                dfs(i);
            }
        }
        printf("%d\n",count);
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;++i){
            if(!vis[i]){
                cnt=0;
                vis[i]=1;
                dfs(i);
                printf("%d",cnt);
                for(int j=1;j<=cnt;++j){
                    printf(" %d",ans[j]);
                }puts("");
            }
        }
    }
    return 0;
}
/* 12 2 3 4 5 6 7 8 9 10 11 12 13 u=2 v=1 u=2 v=3 u=4 v=1 u=4 v=5 u=6 v=5 u=6 v=11 u=8 v=7 u=8 v=9 u=10 v=7 u=10 v=11 u=12 v=3 u=12 v=9 ------------------- 1 12 1 2 3 12 9 8 7 10 11 6 5 4 Process returned 0 (0x0) execution time : 0.148 s Press any key to continue. */