HDU

题意:n个点m条边,找点集个数,点集满足有任意三个点成环,或者三个点互不相连

题解:暴力复杂度O(n^5/120*O(ok))==O(能过)

//#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 mk make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#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 pli pair<ll,int>
#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-12;
const int N=50+10,maxn=60000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;

int ma[N][N];
ll c[N][N];
void init()
{
    c[1][0]=c[1][1]=1;
    for(int i=2;i<N;i++)
    {
        for(int j=0;j<=i;j++)
        {
            if(j==0)c[i][j]=1;
            else c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
        }
    }
}
ll quick(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1)ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}
bool ok(int a,int b,int c)
{
    if(ma[a][b]&&ma[b][c]&&ma[c][a])return 1;
    if(!ma[a][b]&&!ma[b][c]&&!ma[c][a])return 1;
    return 0;
}
bool ok(int a,int b,int c,int d)
{
    if(ok(a, b, c))return 1;
    if(ok(a, b, d)) return 1;
    if(ok(a, c, d))return 1;
    if(ok(b, c, d))return 1;
    return 0;
}
bool ok(int a,int b,int c,int d,int e)
{
    if(ok(a,b,c,d))return 1;
    if(ok(a,b,c,e))return 1;
    if(ok(a,b,e,d))return 1;
    if(ok(a,e,c,d))return 1;
    if(ok(e,b,c,d))return 1;
    return 0;
}
int main()
{
    init();
    int T;scanf("%d",&T);
    for(int _=1;_<=T;_++)
    {
        memset(ma,0,sizeof ma);
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
        {
            int a,b;scanf("%d%d",&a,&b);
            ma[a][b]=ma[b][a]=1;
        }
        ll ans=0;
        if(n>=6)
        {
            ans=quick(2,n);
            for(int i=0;i<=5;i++)
            {
                ans-=c[n][i];
                ans=(ans+mod)%mod;
            }
        }
        for(int i=1;i<=n;i++)
            for(int j=1+i;j<=n;j++)
                for(int k=1+j;k<=n;k++)
                    if(ok(i,j,k))
                        ans++;
        for(int i=1;i<=n;i++)
            for(int j=1+i;j<=n;j++)
                for(int k=1+j;k<=n;k++)
                    for(int u=1+k;u<=n;u++)
                        if(ok(i,j,k,u))
                            ans++;
        for(int i=1;i<=n;i++)
            for(int j=1+i;j<=n;j++)
                for(int k=1+j;k<=n;k++)
                    for(int u=1+k;u<=n;u++)
                        for(int v=1+u;v<=n;v++)
                            if(ok(i,j,k,u,v))
                                ans++;
        printf("Case #%d: %lld
",_,ans%mod);
    }
    return 0;
}

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