제출 #1170204

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1170204		mychecksedad	Reconstruction Project (JOI22_reconstruction)	C++17	70 / 100	5100 ms	462284 KiB

reconstruction

/* Author : Mychecksdead  */
#pragma GCC optimize("unroll-loops,O3")
#pragma GCC target("avx2")
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define MOD (1000000000+7)
#define MOD1 (998244353)
#define pb push_back
#define all(x) x.begin(), x.end()
#define en cout << '\n'
#define ff first
#define ss second
#define pii pair<int,int>
#define vi vector<int>
const int N = 1e6+100, M = 1e5+10, K = 52, MX = 30;

int s[N], p[N];
struct Dsu{
  Dsu(int n){
    for(int j = 0; j <= n; ++j) s[j] = 1, p[j] = j;
  }
  int find(int v){
    if(p[v] == v) return v;
    return p[v] = find(p[v]);
  }
  bool merge(int u, int v){
    u = find(u);
    v = find(v);
    if(u != v){
      if(s[u] > s[v] )swap(u, v);
      p[u] = v;
      s[v] += s[u];
      return true;
    }
    return false;
  }
  bool merge(array<int, 3> e){
    int u = e[1], v = e[2];
    u = find(u);
    v = find(v);
    if(u != v){
      if(s[u] > s[v] )swap(u, v);
      p[u] = v;
      s[v] += s[u];
      return true;
    }
    return false;
  }
};


int n, m, q, maxL[N], maxR[N];
array<int, 3> E[N];
vi L[N], R[N];
void solve(){
  cin >> n >> m;
  for(int j=0;j<m;++j) cin >> E[j][1] >> E[j][2] >> E[j][0];
  E[m] = {-MOD, 0, 0};
  E[m + 1] = {MOD+MOD, 0, 0};
  sort(E, E+m+2, [&](const array<int, 3> &x, const array<int, 3> &y){
    return x[0] < y[0];
  });

  for(int i = 1; i <= m; ++i){
    maxL[i] = maxR[i] = i;
  }
  L[1].pb(0);
  L[1].pb(1);
  L[1].pb(m + 1);
  for(int i = 2; i <= m; ++i){
    Dsu d(n);
    d.merge(E[i]);
    L[i].pb(i);
    for(int j = int(L[i - 1].size())-2; j > 0; --j){
      if(d.merge(E[L[i - 1][j]])) L[i].pb(L[i - 1][j]);
    }
    L[i].pb(0);
    reverse(all(L[i]));
    L[i].pb(m+1);
  }
  R[m].pb(0);
  R[m].pb(m);
  R[m].pb(m + 1);
  for(int i = m - 1; i >= 1; --i){
    Dsu d(n);
    d.merge(E[i]);
    R[i].pb(0);
    R[i].pb(i);
    for(int j = 1; j + 1 < R[i + 1].size(); ++j){
      if(d.merge(E[R[i + 1][j]])) R[i].pb(R[i + 1][j]);
    }
    R[i].pb(m+1);
  }
  for(int i = 1; i <= m; ++i){
    for(int x: L[i]) maxR[x] = max(maxR[x], i);
    for(int x: R[i]) maxL[x] = min(maxL[x], i);
  }
  cin >> q;

  for(int i = 1; i <= q; ++i){
    ll x; cin >> x;
    int pos = upper_bound(E+1, E+m+1, array<int, 3>{(int)x, -1, -1}) - E;
    --pos;
    if(pos == 0){
      ll ans = 0;
      for(int j = 1; j + 1 < R[pos + 1].size(); ++j) ans += E[R[pos + 1][j]][0] - x;
      cout << ans << '\n';
      continue;
    }
    if(pos == m){
      ll ans = 0;
      for(int j = 1; j + 1 < L[pos].size(); ++j) ans += x - E[L[pos][j]][0];
      cout << ans << '\n';
      continue;
    }
    int l = int(L[pos].size()) - 2;
    int r = 1;
    int sz = 1;
    ll ans = 0;
    while(sz < n){
      if(x - E[L[pos][l]][0] <= E[R[pos + 1][r]][0] - x){
        if(maxR[L[pos][l]] >= R[pos + 1][r - 1]){
          ans += abs(E[L[pos][l]][0] - x);
          ++sz;
        }
        --l;
      }else{
        if(maxL[R[pos + 1][r]] <= L[pos][l + 1]){
          ans += abs(E[R[pos + 1][r]][0] - x);
          ++sz;
        }
        ++r;
      }
    }
    cout << ans << '\n';
  }
}


int main(){
  cin.tie(0); ios::sync_with_stdio(0);
  int tt = 1, aa;
  // freopen("in.txt", "r", stdin);
  // freopen("out.txt", "w", stdout);
  while(tt--){
    solve();
    en;
  }
  cerr<<"time taken : "<<(float)clock()/CLOCKS_PER_SEC<<" seconds\n";
  return 0;
} 

• Subtask 1: 3 / 3
• Subtask 2: 4 / 4
• Subtask 3: 0 / 7
• Subtask 4: 28 / 28
• Subtask 5: 35 / 35
• Subtask 6: 0 / 23