Submission #290279

#	Time	Username	Problem	Language	Result	Execution time	Memory
290279		NaimSS	Bridges (APIO19_bridges)	C++14	73 / 100	3077 ms	11296 KiB

bridges

#include <bits/stdc++.h>
#define ld long double
#define endl "\n"
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb(x) push_back(x)
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define per(i, a, b) for(int i = b-1; i>=a ; i--)
#define trav(a, x) for(auto& a : x)
#define allin(a , x) for(auto a : x)
#define Unique(v) sort(all(v));v.erase(unique(all(v)),v.end());
#define sz(v) (int)v.size()
//#define int long long
using namespace std;
typedef vector<int> vi;
#define y1 abacaba
#define left sefude
#define db(x) cerr << #x <<" == "<<x << endl;
#define db2(x,y) cerr<<#x <<" == "<<x<<", "<<#y<<" == "<<y<<endl;
#define db3(x,y,z) cerr << #x<<" == "<<x<<", "<<#y<<" == "<<y<<", "<<#z<<" == "<<z<<endl; 
typedef long long ll;
typedef pair<int,int> pii;
inline ll mod(ll n, ll m ){ ll ret = n%m; if(ret < 0) ret += m; return ret; }
ll gcd(ll a, ll b){return (b == 0LL ? a : gcd(b, a%b));}
ll exp(ll a,ll b,ll m){
    if(b==0LL) return 1LL;
    if(b==1LL) return mod(a,m);
    ll k = mod(exp(a,b/2,m),m);
    if(b&1LL){
        return mod(a*mod(k*k,m),m);
    }
    else return mod(k*k,m);
}

struct DSU{
  vi pai,ps;
  DSU(){}
  DSU(int n){
    pai.resize(n+1);
    ps.resize(n+1);
    for(int i=1;i<=n;i++)pai[i]=i,ps[i]=1;
  }
  int f(int x){
    return pai[x] = (pai[x]==x?x:f(pai[x]));
  }
  int getsz(int x){
    return ps[f(x)];
  }
  void join(int a,int b){
    a=f(a),b=f(b);
    if(a==b)return;
    if(ps[a]>ps[b]){int c=a;a=b,b=c;}
    ps[b]+=ps[a];
    pai[a]=b;
  }
};
const int N = 50100,M = 100100;
const int sq = 400;
vi g[N]; // guarda os ids
int a[M],b[M],c[M];
int res[M];
int vis[N];
int T=0;

int t[M],s[M],w[M];
int m,n;
vector<int> ar[N];
bool jf[M];
DSU dsu(0);
int dfs(int v){
  vis[v] = T;
  int r = dsu.getsz(v);
  for(int to : ar[v])if(vis[to]!=T){
    r+=dfs(to);
  }
  return r;
}
bool used[M];
vi edge;
// O(N*sq^2 + MlogM * M/sq + Q*sq + )
void go(vi qry){
  vi Q,upd;
  

  for(int id : qry){
    if(t[id] == 2)Q.pb(id);
    else{
        upd.pb(id);
        used[s[id]] = 1;
    }
  }
  vi unused;
  for(int i=1;i<=m;i++){ // O(M)
    if(used[i]){
      unused.pb(i);
      continue;
    }
  }
  sort(all(Q),[&](int a,int b){
    return w[a] > w[b];
  }); // sqrt * log(sqrt);
  dsu = DSU(n); // O(N * alpha)
  reverse(all(upd));
  int curedge = 0;
  for(int id : Q){
    while(curedge < sz(edge) and c[edge[curedge]]>=w[id]){ // O(M)
      // gogo
      int ed = edge[curedge];
      if(used[ed]){
        curedge++;
        continue;
      }
      dsu.join(a[ed],b[ed]);
      curedge++;
    }
    T++;
    for(int id2 : upd){
      if(id2 > id || jf[s[id2]])continue;
      jf[s[id2]] = 1;
      if(w[id2] >= w[id]){
        // add edge:
        int ed = s[id2];
     
        ar[dsu.f(a[ed])].pb(dsu.f(b[ed]));
        ar[dsu.f(b[ed])].pb(dsu.f(a[ed]));
      }
    }
    for(int id2 : unused){
      if(!jf[id2] and c[id2] >= w[id]){
        ar[dsu.f(a[id2])].pb(dsu.f(b[id2]));
        ar[dsu.f(b[id2])].pb(dsu.f(a[id2]));
        jf[id2] = 1;
      }
    }
    res[id] = dfs(dsu.f(s[id]));
    // rool back
    for(int id2 : upd){
      jf[s[id2]] = 0;
      int ed = s[id2];
      ar[dsu.f(a[ed])].clear();
      ar[dsu.f(b[ed])].clear();
    }
    for(int id2 : unused){
      jf[id2] = 0;
      ar[dsu.f(a[id2])].clear();
      ar[dsu.f(b[id2])].clear();
    }
  }
  for(int ed : unused){
    used[ed] = 0;
  }

  // update the edges:

  for(int id : qry){
    if(t[id] == 1){
      c[s[id]] = w[id];
    }
  }
  sort(all(edge),[&](int a,int b){
    return c[a] > c[b];
  });
}

int32_t main(){
  fastio;

  cin>>n>>m;
  
  for(int i=1;i<=m;i++){
    cin >> a[i] >> b[i] >> c[i];
    edge.pb(i);
  }
  sort(all(edge),[&](int a,int b){
    return c[a] > c[b];
  });
  int q;
  cin >> q;
  vi qry;
  for(int i=0;i<q;i++){
    cin >> t[i] >> s[i] >> w[i];
    qry.pb(i);
    if(sz(qry) == sq){
      go(qry);
      qry.clear();
    }
  }
  go(qry);
  for(int i=0;i<q;i++){
    if(t[i] == 2)cout << res[i] << endl;
  }
  // math -> gcd it all
  // Did u check N=1? Did you switch N,M?
}

• Subtask 1: 13 / 13
• Subtask 2: 16 / 16
• Subtask 3: 17 / 17
• Subtask 4: 14 / 14
• Subtask 5: 13 / 13
• Subtask 6: 0 / 27