版权声明:写了自己看的,看不懂不能怪我emmmm。 https://blog.csdn.net/qq_40858062/article/details/83999523
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5917
思路:Ramsey定理的通俗表述: 6 个人中至少存在3人相互认识或者相互不认识。所以对于大于6的子集都是满足条件的,组合数算一下就好,对于小于6的子集,最多是n的5次方,直接暴力。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
ll c[58][58],n,m,ans,ma[58][58];
void init()
{
c[0][0]=1;
FOR(i,1,50)
FOR(j,0,i)
{
if(j==0||j==i) c[i][j]=1;
else c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
}
}
bool as(int x,int y,int z)
{
int s=0;
s+=ma[x][y];
s+=ma[x][z];
s+=ma[z][y];
return (s==3||s==0)?1:0;
}
int main()
{
cin.tie(0);
cout.tie(0);
int t,x,y,res=1;
cin>>t;
init();
while(t--)
{
sl(n),sl(m);
ans=0;REW(ma,0);
if(n>=6) FOR(i,6,n) ans=(ans+c[n][i])%mod;
FOR(i,1,m)
{
si(x),si(y);
ma[x][y]=ma[y][x]=1;
}
if(n>=3)
{
FOR(i,1,n)
FOR(j,i+1,n)
FOR(k,j+1,n)
{
if(as(i,j,k)) ans++;
}
}
if(n>=4)
{
FOR(i,1,n)
FOR(j,i+1,n)
FOR(k,j+1,n)
FOR(kk,k+1,n)
{
if(as(i,j,k)||as(i,j,kk)||as(i,k,kk)||as(k,j,kk)) ans++;
}
}
if(n>=5)
{
FOR(i,1,n)
FOR(j,i+1,n)
FOR(k,j+1,n)
FOR(kk,k+1,n)
FOR(kkk,kk+1,n)
{
if(as(i,j,k)||as(i,j,kk)||as(i,j,kkk)||as(i,k,kk)||as(j,kk,kkk)
||as(i,k,kkk)||as(i,kk,kkk)||as(j,k,kk)||as(j,k,kkk)||as(kkk,k,kk)) ans++;
}
}
printf("Case #%d: %lld\n",res++,ans%mod);
}
return 0;
}