Submission #1213208

#	Time	Username	Problem	Language	Result	Execution time	Memory
1213208		mychecksedad	Garden (JOI23_garden)	C++20	0 / 100	682 ms	46744 KiB

garden

/* Author : Mychecksdead  */
#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 n, m, D;
array<int, 2> a[N], b[N];
multiset<int> Y, dif; // Y's of type A's and differences

void add(int y){
  auto it = Y.lower_bound(y);
  if(it != Y.begin()){
    dif.insert(y - (*prev(it)));
  }
  if(it != Y.end()){
    dif.insert((*it) - y);
  }
  if(it != Y.begin() && it != Y.end()){
    dif.erase(dif.find((*it) - (*prev(it))));
  }
  Y.insert(y);
}
void del(int y){
  auto it = Y.lower_bound(y);
  if(it != Y.begin()){
    dif.erase(dif.find(y - (*prev(it))));
  }
  if(next(it) != Y.end()){
    dif.erase(dif.find((*next(it)) - y));
  }
  if(it != Y.begin() && next(it) != Y.end()){
    dif.insert((*next(it)) - (*prev(it)));
  }
  Y.erase(it);
}

int get_max(){
  return max((*prev(dif.end())), (*Y.begin()) + D - (*prev(Y.end()))) - 1; // 2 constant factor gone 
}

void solve(){
  cin >> n >> m >> D;
  for(int i = 1; i <= n; ++i){
    cin >> a[i][0] >> a[i][1];
  }
  for(int i = 1; i <= m; ++i){
    cin >> b[i][0] >> b[i][1];
  }

  sort(a+1, a+1+n);
  sort(b+1, b+1+m);

  for(int i = 1; i <= n; ++i){
    if(i == 1 || a[i - 1][1] != a[i][1]){
      add(a[i][1]);
    }
  }

  int R = a[n][0];
  int ptr = 1;

  int ans = D * D;

  int ptr2 = 1;
  for(int l = 0; l < D; ++l){
    while(ptr <= n && l > a[ptr][0]){
      R = a[ptr][0] + D;
      ++ptr;
    }
    int r = R; // minimum right border basically
     
    // cerr << l << ' ' << r << ":\n";

    while(ptr2 <= m && l > b[ptr2][0]){
      b[ptr2 + m] = {b[ptr2][0] + D, b[ptr2][1]};
      ++ptr2;
    }

    // for(int i = ptr2; i < ptr2 + m; ++i) cerr << b[i][0] << ' ' << b[i][1] << '\n';
    // cerr << '\n';

    // we will try all type B's...

    for(int i = ptr2; i < m + ptr2; ++i){
      if(b[i][0] > r) add(b[i][1]); // let's assume we're taking them all horizontal
    }
    ans = min(ans, (D - get_max()) * (r - l + 1));
    // cerr << "ans: " << (D - get_max()) * (r - l + 1) << '\n';

    for(int i = ptr2; i < ptr2 + m; ++i){
      if(b[i][0] > r){
        del(b[i][1]); // erase em' one by one
      }
      ans = min(ans, (D - get_max()) * (max(r, b[i][0]) - l + 1));
    }
  }
  cout << ans;
}


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: 0 / 15
• Subtask 2: 0 / 6
• Subtask 3: 0 / 8
• Subtask 4: 0 / 16
• Subtask 5: 0 / 30
• Subtask 6: 0 / 25