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(""); } } } }