Codeforces Round #428 (Div. 2)E. Mother of Dragons

http://codeforces.com/contest/839/problem/E

最大团裸题= =，用Bron–Kerbosch算法，复杂度大多博客上没有，维基上查了查大约是O(3^n/3)

最大团： V中取K个顶点，两点间相互连接

最大独立集： V中取K个顶点，两点间不连接

最大团数量 = 补图中最大独立集数

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)

using namespace std;

const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;

int ans,cnt[N],group[N],vis[N];
int n,a[N][N];
bool dfs(int u,int pos)
{
    printf("%d---%d----%d\n",u,pos,ans);
    for(int i=u+;i<=n;i++)
    {
        printf("%d###%d\n",cnt[i],i);
        if(cnt[i]+pos<=ans)return ;
        if(a[u][i])
        {
            int j;
            for(j=;j<pos;j++)if(!a[i][vis[j]])break;
            printf("%d++++\n",j);
            if(j==pos)
            {
                vis[pos]=i;
                if(dfs(i,pos+))return ;
            }
        }
    }
    if(pos>ans)
    {
        for(int i=;i<pos;i++)group[i]=vis[i];
        ans=pos;
        return ;
    }
    return ;
}
void maxclique()
{
    ans=-;
    for(int i=n;i;i--)
    {
        vis[]=i;
        dfs(i,);
        cnt[i]=ans;
        printf("%d!!!\n",cnt[i]);
    }
}
int main()
{
    int k;
    scanf("%d%d",&n,&k);
    for(int i=;i<=n;i++)
        for(int j=;j<=n;j++)
            scanf("%d",&a[i][j]);
    maxclique();
//    printf("%d\n",ans);
    printf("%.12f\n",1.0*k/ans*k/ans*ans*(ans-)/);
    return ;
}
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