题解

我居然都没反应过来二分图内选集合两两不能有边是最大独立集了

我退役吧
显然连边只能在奇数和偶数之间，然后二分图求最大独立集是节点数-最大匹配数

啊当然还有对于1的话只能留一个1

代码

#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define MAXN 3005
#define mo 99994711
#define pb push_back
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef unsigned long long u64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 - '0' + c;
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
int N;
int a[MAXN],matc[MAXN];
bool isprime[1000005],vis[MAXN];
int prime[100005],tot;
struct node {
    int to,next;
}E[MAXN * MAXN / 4];
int head[MAXN],sumE;
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
bool match(int u) {
    for(int i = head[u] ; i ; i = E[i].next) {
        int v = E[i].to;
        if(!vis[v]) {
            vis[v] = 1;
            if(!matc[v] || match(matc[v])) {
                matc[v] = u;
                return true;
            }
        }
    }
    return false;
}
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    sort(a + 1,a + N + 1);
    int p = 1;
    while(p <= N) {
        if(a[p] != 1) break;
        ++p;
    }
    if(a[1] == 1) --p;
    for(int i = p ; i <= N ; ++i) a[i - p + 1] = a[i];
    N = N - p + 1;
    for(int i = 1 ; i <= N ; ++i) {
        if(a[i] % 2 == 0) {
            for(int j = 1 ; j <= N ; ++j) {
                if(isprime[a[i] + a[j]]) add(i,j);
            }
        }
    }
    int ans = N;
    for(int i = 1 ; i <= N ; ++i) {
        if(a[i] % 2 == 0) {
            memset(vis,0,sizeof(vis));
            if(match(i)) --ans;
        }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    memset(isprime,1,sizeof(isprime));
    for(int i = 2 ; i <= 1000000 ; ++i) {
        if(isprime[i]) {
            prime[++tot] = i;
        }
        for(int j = 1 ; j <= tot ; ++j) {
            if(prime[j] > 1000000 / i) break;
            isprime[i * prime[j]] = 0;
            if(i % prime[j] == 0) break;
        }
    }
    Solve();
}