【二分图最大匹配】【匈牙利算法】zoj3988 Prime Set

#include<cstdio>
#include<algorithm>
#include<cstring>
typedef long long ll;
using namespace std;
int T,n,K,a[3005];
bool notprime[2000005];
bool cmp(const int &a,const int &b){
    return a>b;
}
int e,first[3005],next[3005*3005],v[3005*3005];
void AddEdge(int U,int V){
    v[++e]=V;
    next[e]=first[U];
    first[U]=e;
}
int mat[3005],yi;
bool vis[3005];
bool dfs(int U)
{
    for(int i=first[U];i;i=next[i]){
        if(!vis[v[i]]){
            vis[v[i]]=1;
            if(mat[v[i]]==-1 || dfs(mat[v[i]])){
                mat[v[i]]=U;
                return 1;
            }
        }
    }
    return 0;
}
int ru[3005],ru1[3005],right1,right2;
int main(){
    notprime[1]=1;
    for(int i=2;i<=2000000;++i){
        for(ll j=(ll)i*(ll)i;j<=2000000ll;j+=(ll)i){
            notprime[j]=1;
        }
    }
    scanf("%d",&T);
    for(;T;--T){
        e=0;
        memset(mat,-1,sizeof(mat));
        memset(first,0,sizeof(first));
        memset(ru,0,sizeof(ru));
        memset(ru1,0,sizeof(ru1));
        scanf("%d%d",&n,&K);
        yi=0;
        for(int i=1;i<=n;++i){
            scanf("%d",&a[i]);
            if(a[i]==1){
                ++yi;
            }
        }
        sort(a+1,a+n+1,cmp);
        for(int i=1;i<=n;++i){
            for(int j=i+1;j<=n;++j){
                if(!(a[i]==1 && a[j]==1) && !notprime[a[i]+a[j]]){
                    if(a[i]&1){
                        AddEdge(i,j);
                        ++ru[j];
                        if(a[i]==1){
                            ++ru1[j];
                        }
                    }
                    else{
                        AddEdge(j,i);
                        ++ru[i];
                        if(a[j]==1){
                            ++ru1[i];
                        }
                    }
                }
            }
        }
        bool tp=(yi==1 ? 1 : 0);
        right1=right2=0;
        for(int i=1;i<=n;++i){
            if(!(a[i]&1)){
                if(ru[i]>ru1[i]){
                    ++right1;
                }
                if(ru[i]){
                    ++right2;
                }
                if(ru1[i]){
                    tp=0;
                }
            }
        }
        if(tp){
            yi=0;
        }
        int sum=0,cnt=0,nowleft=0,ans=0;
        for(int i=1;i<=n;++i){
            if((a[i]&1)){
                ++nowleft;
                if(a[i]!=1 && first[i]){
                    memset(vis,0,sizeof(vis));
                    if(dfs(i)){
                        ++sum;
                    }
                }
                else if(a[i]==1 && !tp){
                    if(!cnt){
                        if(K<=sum+yi/2){
                            ans=max(ans,2*K);
                        }
                        else{
                            ans=max(ans,2*sum+yi+min(K-sum-yi/2-yi%2,nowleft-1+right1-sum*2));
                        }
                    }
                    ++cnt;
                    memset(vis,0,sizeof(vis));
                    if(dfs(i)){
                        ++sum;
                    }
                    if(K<=sum+(yi-cnt)/2){
                        ans=max(ans,2*K);
                    }
                    else{
                        ans=max(ans,2*sum+(yi-cnt)+min(K-sum-(yi-cnt)/2-(yi-cnt)%2,nowleft+right2-sum*2));
                    }
                }
            }
        }
        if(!cnt){
            if(K<=sum){
                ans=max(ans,2*K);
            }
            else{
                ans=max(ans,2*sum+min(K-sum,nowleft+right1-sum*2));
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}