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=100000,MAXM=100000,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=0; memset(G,-1,sizeof(G[0])*N);
    }
    void link(int u,int v,int c)
    {
        E[sz]=Edge(v,c,0,G[u]); G[u]=sz++;
        E[sz]=Edge(u,0,0,G[v]); G[v]=sz++;
    }
    int ISAP(int S,int T)
    {//S -> T
        int maxflow=0,aug=inf,flag=false,u,v;
        for (int i=0;i<N;++i)cur[i]=G[i],gap[i]=dis[i]=0;
        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]+1)
                {
                    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]^1].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]])==0) break;
            ++gap[dis[u]=mx+1]; u=pre[u];
        }
        return maxflow;
    }
    bool bfs(int S,int T)
    {
        static int Q[MAXN]; memset(dis,-1,sizeof(dis[0])*N);
        dis[S]=0; Q[0]=S;
        for (int h=0,t=1,u,v,it;h<t;++h)
        {
            for (u=Q[h],it=G[u];~it;it=E[it].nx)
            {
                if (dis[v=E[it].v]==-1&&E[it].c>E[it].f)
                {
                    dis[v]=dis[u]+1; Q[t++]=v;
                }
            }
        }
        return dis[T]!=-1;
    }
    int dfs(int u,int T,int low)
    {
        if (u==T) return low;
        int ret=0,tmp,v;
        for (int &it=cur[u];~it&&ret<low;it=E[it].nx)
        {
            if (dis[v=E[it].v]==dis[u]+1&&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^1].f-=tmp;
                }
            }
        }
        if (!ret) dis[u]=-1; return ret;
    }
    int dinic(int S,int T)
    {
        int maxflow=0,tmp;
        while (bfs(S,T))
        {
            memcpy(cur,G,sizeof(G[0])*N);
            while (tmp=dfs(S,T,inf)) maxflow+=tmp;
        }
        return maxflow;
    }
}
using namespace NetFlow;


int One;
vector<int> Even,Odd;
const int MAXN_ = 10000;
bool _flag[MAXN_];
int _primes[MAXN_], _pi;
void GetPrime_1()
{
    int i, j;
    _pi = 0;
    memset(_flag, false, sizeof(_flag));
    for (i = 2; i < MAXN_; i++)
        if (!_flag[i])
        {
            _primes[i] = 1;//素数标识为1
            for (j = i; j < MAXN_; j += i)
                _flag[j] = true;
        }
}
void Clear()
{
    Even.clear();
    Odd.clear();
    init(100000);
    One=0;
}
bool work()
{
    int n;scanf("%d",&n);
    int flag = 0;
    for(int i=1;i<=n;i++)
    {
        int x;scanf("%d",&x);
        if(x==1)One++;
        else if(x%2==1)Odd.push_back(x);
        else Even.push_back(x);
    }
    if(Odd.size()>Even.size())return 0;
    if(Odd.size()+One<Even.size())return 0;
    int In = 0;
    while(Odd.size()<Even.size())
    {
        flag = 1;
        Odd.push_back(1);
        One--;
        In++;
    }
    if(flag==0&&(One==1||One==2))return 0;
    int tmp = 0;
    for(int i=0;i<Odd.size();i++)
        link(0,1+i,2);
    for(int i=0;i<Even.size();i++)
        link(n+2+i,600,2);
    for(int i=0;i<Odd.size();i++)
    {
        for(int j=0;j<Even.size();j++)
        {
            if(_primes[Odd[i]+Even[j]])
            {
                if(Odd[i]==1&&One>0&&In>0)
                    link(1+i,n+2+j,2);
                else
                    link(1+i,n+2+j,1);
            }
        }
    }
    if(Even.size()*2==dinic(0,600))return 1;
    return 0;
}
int main()
{
    GetPrime_1();
    int t;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        Clear();
        printf("%s\n",work()?"YES":"NO");
    }
}