loj526 "LibreOJ β Round # 4" subset

analysis

The largest group seeking to meet the conditions

We can consider building a complementary graph

It revealed between odd and even among

Since it must satisfy the above must be odd Complements a form into a set of even a bipartite graph

Seeking bipartite graph can be set independently of the maximum

Code

#include<bits/stdc++.h>
using namespace std;
#define int long long
int n,m,d[1010],g[1010][1010],a[1010],b[1010],c1,c2,ans,T,used[1010],wh[1010];
inline bool work(int x){
    for(int i=1;i<=m;i++)
      if(g[x][i]&&used[i]!=T){
          used[i]=T;
          if(!wh[i]||work(wh[i])){
            wh[i]=x;
            return 1;
        }
      }
    return 0;
}
inline void go(){
    for(int i=1;i<=n;i++){
      ++T;
      if(work(i))ans++;
    }
}
signed main(){
    int i,j,k;
    scanf("%lld",&n);
    for(i=1;i<=n;i++){
      scanf("%lld",&d[i]);
      if(d[i]&1)a[++c1]=d[i];
        else b[++c2]=d[i];
    }
    n=c1,m=c2;
    for(i=1;i<=c1;i++)
      for(j=1;j<=c2;j++)
        if(__gcd(a[i],b[j])==1&&__gcd(a[i]+1,b[j]+1)==1)g[i][j]=1;
    go();
    printf("%lld",n+m-ans);
    return 0;
}