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-;
const int N=+,maxn=+,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; int ma[N][N];
ll c[N][N];
void init()
{
c[][]=c[][]=;
for(int i=;i<N;i++)
{
for(int j=;j<=i;j++)
{
if(j==)c[i][j]=;
else c[i][j]=(c[i-][j]+c[i-][j-])%mod;
}
}
}
ll quick(ll a,ll b)
{
ll ans=;
while(b)
{
if(b&)ans=ans*a%mod;
a=a*a%mod;
b>>=;
}
return ans;
}
bool ok(int a,int b,int c)
{
if(ma[a][b]&&ma[b][c]&&ma[c][a])return ;
if(!ma[a][b]&&!ma[b][c]&&!ma[c][a])return ;
return ;
}
bool ok(int a,int b,int c,int d)
{
if(ok(a, b, c))return ;
if(ok(a, b, d)) return ;
if(ok(a, c, d))return ;
if(ok(b, c, d))return ;
return ;
}
bool ok(int a,int b,int c,int d,int e)
{
if(ok(a,b,c,d))return ;
if(ok(a,b,c,e))return ;
if(ok(a,b,e,d))return ;
if(ok(a,e,c,d))return ;
if(ok(e,b,c,d))return ;
return ;
}
int main()
{
init();
int T;scanf("%d",&T);
for(int _=;_<=T;_++)
{
memset(ma,,sizeof ma);
int n,m;
scanf("%d%d",&n,&m);
for(int i=;i<m;i++)
{
int a,b;scanf("%d%d",&a,&b);
ma[a][b]=ma[b][a]=;
}
ll ans=;
if(n>=)
{
ans=quick(,n);
for(int i=;i<=;i++)
{
ans-=c[n][i];
ans=(ans+mod)%mod;
}
}
for(int i=;i<=n;i++)
for(int j=+i;j<=n;j++)
for(int k=+j;k<=n;k++)
if(ok(i,j,k))
ans++;
for(int i=;i<=n;i++)
for(int j=+i;j<=n;j++)
for(int k=+j;k<=n;k++)
for(int u=+k;u<=n;u++)
if(ok(i,j,k,u))
ans++;
for(int i=;i<=n;i++)
for(int j=+i;j<=n;j++)
for(int k=+j;k<=n;k++)
for(int u=+k;u<=n;u++)
for(int v=+u;v<=n;v++)
if(ok(i,j,k,u,v))
ans++;
printf("Case #%d: %lld\n",_,ans%mod);
}
return ;
} /***********************
***********************/