BC#65 T5 ZYB's Prime

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

完全不会啊TAT。。 

其实官方题解已经说的很清楚了。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <queue>
#include <cmath>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 405
#define ll long long
#define inf int(1e9)
using namespace std;
int a[maxn],head[maxn],cur[maxn],uu[maxn],odd[maxn],even[maxn];
int tot,t,s;
struct data{int obj,pre,c;
}e[200500];
int read(){
    int x=0,f=1; char ch=getchar();
    while (!isdigit(ch)){ if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
void insert(int x,int y,int z){
    e[++tot].obj=y; e[tot].c=z; e[tot].pre=head[x]; head[x]=tot;
    e[++tot].obj=x; e[tot].c=0; e[tot].pre=head[y]; head[y]=tot;
}
bool bfs(){
    queue<int >q; clr(uu,-1); uu[s]=0; q.push(s);
    while (!q.empty()){
        int u=q.front(); q.pop();
        for (int j=head[u];j;j=e[j].pre){
            int v=e[j].obj;
            if (uu[v]==-1&&e[j].c) uu[v]=uu[u]+1,q.push(v);    
        }
    }
    if (uu[t]==-1) return 0;
    return 1;
}
int dfs(int x,int mx){
    if (x==t) return mx;
    int used=0;
    for (int j=cur[x];j;j=e[j].pre){
        int v=e[j].obj;
        if (uu[v]==uu[x]+1){
            int w=dfs(v,min(e[j].c,mx-used));
            used+=w; e[j].c-=w; e[j^1].c+=w;
            if (e[j].c) cur[x]=j;
            if (used==mx) return mx;
        }
    }
    if (!used) uu[x]=-1;
    return used;
}
int dinic(){
    int ans=0;
    while (bfs()){
        rep(i,s,t) cur[i]=head[i];
        ans+=dfs(s,inf);
    }
    return ans;
}
bool jud(int x){
    int s=int(sqrt(x))+1;
    rep(i,2,s) if (x%i==0) return 0;
    return 1;
}
bool work(){    
    tot=1; clr(head,0); clr(odd,0); clr(even,0); //clr(one,0);
    int odd_cnt=0,even_cnt=0,one_in=0,one_out=0,one_cnt=0;
    int n=read();
    rep(i,1,n) {
        a[i]=read(); 
           if (a[i]>1&&(a[i]&1)) odd[++odd_cnt]=a[i];
        else if (a[i]>1&&((a[i]&1)==0)) even[++even_cnt]=a[i];
        else one_cnt++;    
    }
    int need=even_cnt-odd_cnt;
    rep(i,1,one_cnt){
        if (need) {
               one_in++;
               odd[++odd_cnt]=1;
               need--;
           }
           else one_out++;
    }
    if ((!one_in && (one_out==2||one_out==1)) || need!=0) return 0;
    s=0; t=2*odd_cnt+1; 
    int cnt=odd_cnt;
    rep(i,1,cnt) insert(s,i,2);
    rep(i,1,cnt) insert(i+cnt,t,2);
    rep(i,1,cnt) rep(j,1,cnt){
        if (jud(odd[i]+even[j])) {
            if (odd[i]==1&&one_out>0) insert(i,j+cnt,2);
            else insert(i,j+cnt,1);
        }
    }
    if (dinic()!=2*cnt) return 0;
    return 1;
}
int main(){
    int T=read();
    while (T--){
        if (work()) puts("YES"); else puts("NO");
    }
    return 0;
}
原文地址:https://www.cnblogs.com/ctlchild/p/5023135.html