题意

分析

代码

#include 
    
     #include 
     
      #include 
      
       #define MP make_pair#define PB emplace_back#define fi first#define se second#define ZERO(x) memset((x), 0, sizeof(x))#define ALL(x) (x).begin(),(x).end()#define rep(i, a, b) for (repType i = (a); i <= (b); ++i)#define per(i, a, b) for (repType i = (a); i >= (b); --i)#define QUICKIO                  \    ios::sync_with_stdio(false); \    cin.tie(0);                  \    cout.tie(0);using namespace std;typedef long long ll;typedef int repType;const int MAXN=130;bool mat[MAXN][MAXN];int n,m,ans;int done[MAXN][MAXN], notyet[MAXN][MAXN], searched[MAXN][MAXN];void dfs(int n, int dcnt, int ncnt, int scnt){    if(!ncnt && !scnt) ans++;    if(ans>1000) return;    int key=notyet[n][1];    rep(j,1,ncnt)    {        int v=notyet[n][j], tmp_ncnt=0, tmp_scnt=0;        if(mat[key][v]) continue;        memcpy(done[n+1],done[n],sizeof(int)*(dcnt+1));        done[n+1][dcnt+1]=v;        rep(i,1,ncnt) if(mat[v][notyet[n][i]])            notyet[n+1][++tmp_ncnt]=notyet[n][i];        rep(i,1,scnt) if(mat[v][searched[n][i]])            searched[n+1][++tmp_scnt]=searched[n][i];        dfs(n+1, dcnt+1, tmp_ncnt, tmp_scnt);        notyet[n][j]=0;        searched[n][++scnt]=v;    }}int main(){    while(cin>>n>>m)    {        ZERO(mat); ans=0;        rep(i,1,m)        {            int x,y;            cin>>x>>y;            mat[x][y]=mat[y][x]=true;        }        rep(i,1,n) notyet[1][i]=i;        dfs(1,0,n,0);        if(ans>1000) cout<<"Too many maximal sets of friends."<