随一个序列然后贪心即可。

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#include<bits/stdc++.h>
using namespace std;
int n,a[55],v[55][55],cnt[55],b[55];
vector<int> ans;
mt19937 rd((unsigned long long)new char);
int main(){
    cin>>n;
    for(int i=1;i<=n;++i)b[i]=i;
    int x,y;
    while(cin>>x>>y)v[x][y]=v[y][x]=1,++cnt[x],++cnt[y];
    while(1.0*clock()/CLOCKS_PER_SEC<=0.985){
        vector<int> g;g.push_back(b[1]);
        for(int i=2;i<=n;++i){
            if(cnt[b[i]]<g.size())continue;
            int c=0;
            for(int j:g)c+=v[b[i]][j];
            if(c==g.size())g.push_back(b[i]);
        }
        if(ans.size()<g.size())ans=g;
        if(ans.size()==n)break;
        shuffle(b+1,b+n+1,rd);
    }
    cout<<ans.size()<<'\n';
    return 0;
}