uva247

对名字编号以后建图，然后求传递闭包，两个点在一个圈中的充分条件是d[i][j]=1&&d[j][i]=1，证明略

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iomanip>
#include<assert.h>
#include<ctime>
#include<vector>
#include<list>
#include<map>
#include<set>
#include<sstream>
#include<stack>
#include<queue>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
#define me(s)  memset(s,0,sizeof(s))
#define mp make_pair
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define F first
#define S second
#define pf printf
#define sf scanf
#define Di(x) int x;scanf("%d",&x)
#define in(x) inp(x)
#define in2(x,y) inp(x),inp(y)
#define in3(x,y,z) inp(x),inp(y),inp(z)
#define ins(x) scanf("%s",x)
#define ind(x) scanf("%lf",&x)
#define IO ios_base::sync_with_stdio(0);cin.tie(0)
#define READ freopen("C:/Users/ASUS/Desktop/in.txt","r",stdin)
#define WRITE freopen("C:/Users/ASUS/Desktop/out.txt","w",stdout)
template<class T> void inp(T &x) {//读入优化
    char c = getchar(); x = 0;
    for (; (c < 48 || c>57); c = getchar());
    for (; c > 47 && c < 58; c = getchar()) { x = (x << 1) + (x << 3) + c - 48; }
}
typedef pair <int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-15;
const int maxn=25;
vector<string> names;
map<string,int> ID;
int GET_ID(string s){
    if(ID.count(s)) return ID[s];
    names.push_back(s);
    return ID[s]=names.size()-1;
}
int n,m;
int d[maxn][maxn];
void init(){
    ID.clear();
    names.clear();
    me(d);
    for(int i=0;i<n;i++)
    d[i][i]=1;
}
int vis[maxn];
void dfs(int u){
    vis[u]=1;
    for(int i=0;i<n;i++){
        if(!vis[i]&&d[u][i]&&d[i][u]) {
            pf(", %s",names[i].c_str());
            dfs(i);
        }
    }
}
int main(){
    //READ;
    //WRITE;
    int cas=0;
    while(sf("%d%d",&n,&m)==2&&n){
        
        init();
        string u,v;
        for(int i=0;i<m;i++){
            cin>>u>>v;
            d[GET_ID(u)][GET_ID(v)]=1;
        }
        for(int k=0;k<n;k++){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                d[i][j]|=d[i][k]&d[k][j];
        }}}
        if(cas) puts("");
        pf("Calling circles for data set %d:\n",++cas);
        me(vis);
        for(int i=0;i<n;i++){
            if(!vis[i]){
                pf("%s",names[i].c_str());
                dfs(i);
                puts("");
            }
        }
    }

}