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5938번: Daisy Chains in the Field ↗

Solutions

C++14
2.0 KB | 2029 chars
/*
[5938: Daisy Chains in the Field](https://www.acmicpc.net/problem/5938)

Tier: Silver 3 
Category: bfs, dfs, graphs, graph_traversal
*/

#include <bits/stdc++.h>

using namespace std;

#define for1(s, e) for(int i = s; i < e; i++)
#define for1j(s, e) for(int j = s; j < e; j++)
#define forEach(k) for(auto i : k)
#define forEachj(k) for(auto j : k)
#define sz(vct) vct.size()
#define all(vct) vct.begin(), vct.end()
#define sortv(vct) sort(vct.begin(), vct.end())
#define uniq(vct) sort(all(vct));vct.erase(unique(all(vct)), vct.end())
#define fi first
#define se second
#define INF (1ll << 60ll)

typedef unsigned long long ull;
typedef long long ll;
typedef ll llint;
typedef unsigned int uint;
typedef unsigned long long int ull;
typedef ull ullint;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef pair<double, int> pdi;
typedef pair<string, string> pss;

typedef vector<int> iv1;
typedef vector<iv1> iv2;
typedef vector<ll> llv1;
typedef vector<llv1> llv2;

typedef vector<pii> piiv1;
typedef vector<piiv1> piiv2;
typedef vector<pll> pllv1;
typedef vector<pllv1> pllv2;
typedef vector<pdd> pddv1;
typedef vector<pddv1> pddv2;

const double EPS = 1e-8;
const double PI = acos(-1);

template<typename T>
T sq(T x) { return x * x; }

int sign(ll x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(int x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(double x) { return abs(x) < EPS ? 0 : x < 0 ? -1 : 1; }

bool visited[1100];
vector<int> adj[1100];

void dfs(int u) {
  visited[u] = true;

  forEach(adj[u]) {
    if(!visited[i]) dfs(i);
  }
}

void solve() {
  int n, m;

  cin >> n >> m;

  for1(0, m) {
    int a, b;

    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }

  dfs(1);
  bool chk = true;
  for1(2, n + 1) {
    if(!visited[i]) {
      chk = false;
      cout << i << "\n";
    }
  }
    
  if(chk) cout << 0;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);cout.tie(NULL);
  int tc = 1; // cin >> tc;
  while(tc--) solve();

}