Submission #991906

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Source Code Expand

#include <bits/stdc++.h>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}

using ll=long long;
using R=long double;
const R EPS=1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r){return(r > EPS)-(r < -EPS);}
inline R sq(R x){return sqrt(max(x,0.0L));}

const int dx[8]={1,0,-1,0,1,-1,-1,1};
const int dy[8]={0,1,0,-1,1,1,-1,-1};

// Problem Specific Parameter:
// Description: 素集合を管理するデータ構造 
// TimeComplexity: 初期化$\mathcal{O}(n)$ 更新$\mathcal{O}(\log n)$
// Verifyed: AOJ DSL_1_A

struct Union_find{
	Union_find(int n){par.resize(n),iota(_all(par),0);}
	int find(int x){return (par[x]==x)?x:par[x]=find(par[x]);}
	void unite(int a,int b){a=find(a),b=find(b);par[a]=b;}
	bool same(int a,int b){return find(a)==find(b);}
	vector<int> par;
};

vector<int> lang[100010];

int main(void){
	int n,m;
	cin >> n >> m;

	Union_find uf(n);
	rep(i,n){
		int k;
		cin >> k;
		rep(j,k){
			int in;
			cin >> in,in--;
			lang[in].push_back(i);
		}
	}

	rep(i,m){
		if(lang[i].empty()) continue;
		int lead=lang[i][0];
		for(auto &it:lang[i]) uf.unite(lead,it);
	}

	bool ok=true;
	rep(i,n) if(uf.same(0,i)==false) ok=false;

	if(ok)
		puts("YES");
	else
		puts("NO");
	return 0;
}

Submission Info

Judge Result