제출 #1213658

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1213658		mychecksedad	Garden (JOI23_garden)	C++20	0 / 100	302 ms	19552 KiB

garden

/* Author : Mychecksdead  */
#include<bits/stdc++.h>
// #pragma GCC optimize("O3,unroll-loops")
// #pragma GCC target("avx2,bmi,bmi2")
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>
#define cerr if(false) cerr
const int N = 1e6+100, M = 1e5+10, K = 52, MX = 30;


int n, m, D, mx[N], L[N], R[N];
array<int, 2> a[N], b[N], c[N];
vector<int> cnt(N), V(N, -1);
void add(int y, int idx){
  cnt[y]++;
  V[y] = idx; // only the last one is important
}
void del(int y){
  cnt[y]--;
}

void process(){
  int cur = 0, last = -1, mn = MOD;
  for(int i = 0; i < D; ++i){
    cerr << cnt[i] << ' ';
    if(cnt[i]){
      if(mn == MOD) mn = i;
      if(last != -1){
        cur = max(cur, i - last);
      }
      last = i;
    }
  }
  cerr << '\n';
  cur = max(cur, mn + D - last);
  mx[0] = cur - 1;

  cerr << mn  << ' ' << last<< ' ' << mx[0] << " f\n";
  // mx[i] = pt=1...i is popped, what's the answer

  vector<pii> Q;
  for(int i = 0; i < D; ++i) if(V[i] != -1) Q.pb({i - D, V[i]});
  for(int i = 0; i < D; ++i){
    if(V[i] != -1){ 
      while(Q.size() && Q.back().ss < V[i]) Q.pop_back();
      if(Q.size()){
        L[i] = Q.back().ff;
      }else assert(false);
      Q.pb({i, V[i]});
    }
  }
  Q.clear();
  for(int i = D-1; i >= 0; --i) if(V[i] != -1) Q.pb({i + D, V[i]});
  for(int i = D-1; i >= 0; --i){
    if(V[i] != -1){ 
      while(Q.size() && Q.back().ss < V[i]) Q.pop_back();
      if(Q.size()){
        R[i] = Q.back().ff;
      }else assert(false);
      Q.pb({i, V[i]});
    }
  }
  for(int i = 1; i <= D; ++i) mx[i] = -1;
  for(int i = 0; i < D; ++i) cerr << V[i] << ' '<< L[i] << ' ' << R[i] << '\n';
  for(int i = 0; i < D; ++i) if(V[i] >= 0 && V[i] < MOD) mx[V[i]] = max(mx[V[i]], R[i] - L[i] - 1);
  for(int i = 1; i <= D; ++i) mx[i] = max(mx[i], mx[i - 1]);
}

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


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

  vi A;

  for(int i = 1; i <= n; ++i){
    if(i == 1 || c[i - 1][0] != c[i][0]){
      A.pb(c[i][0]);
    }
  }
  // add(c[1][0] + D, 0);
  // A.pb(c[1][0] + D);

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

  int ans = D * D;

  int ptr2 = 1;

  vector<deque<int>> Y(D); // y position of B's

  for(int i = 1; i <= m; ++i) Y[b[i][1]].pb(b[i][0]); 

  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]){
      Y[b[ptr2][1]].pop_front();
      b[ptr2 + m] = {b[ptr2][0] + D, b[ptr2][1]};
      Y[b[ptr2][1]].pb(b[ptr2][0] + D);
      ++ptr2;
    } 

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

    for(int i = 0, pt = 1; i < D; ++i) if(Y[i].size() && Y[i].back() > r){
      add(i, pt++);
    }

    for(int x: A) add(x, MOD);

    process();

    ans = min(ans, (D - mx[0]) * (r - l + 1));
    cerr << D-mx[0]  << ' ' << (r - l + 1) << " hmm\n";

    cerr << mx[0] << " hell\n";
    
    // this is the TLE area...
    for(int i = 0, pt = 1; i < D; ++i) if(Y[i].size() && Y[i].back() > r){
      ans = min(ans, (D - mx[pt++]) * (max(r, Y[i].back()) - l + 1));
    }

    V.clear();
    V.resize(D, -1);

    for(int i = ptr2; i < ptr2 + m; ++i) if(b[i][0] > r) del(b[i][1]); // currently all B's are horizontal
  }
  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;
} 

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• Subtask 2: 0 / 6
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