题意

分析

代码

#include <iostream>
#include <cstring>
#include <algorithm>
#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."<<endl;
        else cout<<ans<<endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/samhx/p/POJ-2989.html