hdu 5594 ZYB's Prime 最大流

ZYB's Prime

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5594

Description

After getting 600 scores in NOIP,ZYB(ZJ−267) creates a problem:you are given N numbers,now you are asked to divide them into K groups（K≥1),the
number of each group must be no less than 3,and put all the numbers in a group into a ring,the sum of every two adjacent numbers must be a prime.ZYB want to ask
you whether the N numbers can be divided or not?

Input

In the first line there is the testcase T.

For each teatcase:

In the first line there is one number N.
In the next line there are N numbers Ai.

1≤T≤50,1≤N≤200,1≤Ai≤200,for 60% cases N≤20.

Output

For each testcase,print the YES or NO.

Sample Input

2
7
3 4 8 9 1 1 1
3
1 2 3

Sample Output

YES
NO

HINT

题意

题解：

代码：

#include<iostream>
#include<stdio.h>
#include<vector>
#include<cstring>
using namespace std;

namespace NetFlow
{
    const int MAXN=,MAXM=,inf=1e9;
    struct Edge
    {
        int v,c,f,nx;
        Edge() {}
        Edge(int v,int c,int f,int nx):v(v),c(c),f(f),nx(nx) {}
    } E[MAXM];
    int G[MAXN],cur[MAXN],pre[MAXN],dis[MAXN],gap[MAXN],N,sz;
    void init(int _n)
    {
        N=_n,sz=; memset(G,-,sizeof(G[])*N);
    }
    void link(int u,int v,int c)
    {
        E[sz]=Edge(v,c,,G[u]); G[u]=sz++;
        E[sz]=Edge(u,,,G[v]); G[v]=sz++;
    }
    int ISAP(int S,int T)
    {//S -> T
        int maxflow=,aug=inf,flag=false,u,v;
        for (int i=;i<N;++i)cur[i]=G[i],gap[i]=dis[i]=;
        for (gap[S]=N,u=pre[S]=S;dis[S]<N;flag=false)
        {
            for (int &it=cur[u];~it;it=E[it].nx)
            {
                if (E[it].c>E[it].f&&dis[u]==dis[v=E[it].v]+)
                {
                    if (aug>E[it].c-E[it].f) aug=E[it].c-E[it].f;
                    pre[v]=u,u=v; flag=true;
                    if (u==T)
                    {
                        for (maxflow+=aug;u!=S;)
                        {
                            E[cur[u=pre[u]]].f+=aug;
                            E[cur[u]^].f-=aug;
                        }
                        aug=inf;
                    }
                    break;
                }
            }
            if (flag) continue;
            int mx=N;
            for (int it=G[u];~it;it=E[it].nx)
            {
                if (E[it].c>E[it].f&&dis[E[it].v]<mx)
                {
                    mx=dis[E[it].v]; cur[u]=it;
                }
            }
            if ((--gap[dis[u]])==) break;
            ++gap[dis[u]=mx+]; u=pre[u];
        }
        return maxflow;
    }
    bool bfs(int S,int T)
    {
        static int Q[MAXN]; memset(dis,-,sizeof(dis[])*N);
        dis[S]=; Q[]=S;
        for (int h=,t=,u,v,it;h<t;++h)
        {
            for (u=Q[h],it=G[u];~it;it=E[it].nx)
            {
                if (dis[v=E[it].v]==-&&E[it].c>E[it].f)
                {
                    dis[v]=dis[u]+; Q[t++]=v;
                }
            }
        }
        return dis[T]!=-;
    }
    int dfs(int u,int T,int low)
    {
        if (u==T) return low;
        int ret=,tmp,v;
        for (int &it=cur[u];~it&&ret<low;it=E[it].nx)
        {
            if (dis[v=E[it].v]==dis[u]+&&E[it].c>E[it].f)
            {
                if (tmp=dfs(v,T,min(low-ret,E[it].c-E[it].f)))
                {
                    ret+=tmp; E[it].f+=tmp; E[it^].f-=tmp;
                }
            }
        }
        if (!ret) dis[u]=-; return ret;
    }
    int dinic(int S,int T)
    {
        int maxflow=,tmp;
        while (bfs(S,T))
        {
            memcpy(cur,G,sizeof(G[])*N);
            while (tmp=dfs(S,T,inf)) maxflow+=tmp;
        }
        return maxflow;
    }
}
using namespace NetFlow;

int One;
vector<int> Even,Odd;
const int MAXN_ = ;
bool _flag[MAXN_];
int _primes[MAXN_], _pi;
void GetPrime_1()
{
    int i, j;
    _pi = ;
    memset(_flag, false, sizeof(_flag));
    for (i = ; i < MAXN_; i++)
        if (!_flag[i])
        {
            _primes[i] = ;//素数标识为1
            for (j = i; j < MAXN_; j += i)
                _flag[j] = true;
        }
}
void Clear()
{
    Even.clear();
    Odd.clear();
    init();
    One=;
}
bool work()
{
    int n;scanf("%d",&n);
    int flag = ;
    for(int i=;i<=n;i++)
    {
        int x;scanf("%d",&x);
        if(x==)One++;
        else if(x%==)Odd.push_back(x);
        else Even.push_back(x);
    }
    if(Odd.size()>Even.size())return ;
    if(Odd.size()+One<Even.size())return ;
    int In = ;
    while(Odd.size()<Even.size())
    {
        flag = ;
        Odd.push_back();
        One--;
        In++;
    }
    if(flag==&&(One==||One==))return ;
    int tmp = ;
    for(int i=;i<Odd.size();i++)
        link(,+i,);
    for(int i=;i<Even.size();i++)
        link(n++i,,);
    for(int i=;i<Odd.size();i++)
    {
        for(int j=;j<Even.size();j++)
        {
            if(_primes[Odd[i]+Even[j]])
            {
                if(Odd[i]==&&One>&&In>)
                    link(+i,n++j,);
                else
                    link(+i,n++j,);
            }
        }
    }
    if(Even.size()*==dinic(,))return ;
    return ;
}
int main()
{
    GetPrime_1();
    int t;
    scanf("%d",&t);
    for(int cas=;cas<=t;cas++)
    {
        Clear();
        printf("%s\n",work()?"YES":"NO");
    }
}