Submission #999940

#	Time	Username	Problem	Language	Result	Execution time	Memory
999940		AdamGS	Cultivation (JOI17_cultivation)	C++17	55 / 100	2015 ms	2844 KiB

cultivation

#include<bits/stdc++.h>
using namespace std;
typedef long double ld;
typedef long long ll;
#define rep(a, b) for(int a = 0; a < (b); ++a)
#define st first
#define nd second
#define pb push_back
#define all(a) a.begin(), a.end()
const ll INF=2e9+7;
const int LIM=307;
pair<ll,ll>T[LIM];
ll tmi[LIM][LIM], tma[LIM][LIM], R, C, n, akt1, akt2, akt3;
vector<pair<ll,ll>>Ci, Di, Ei;
ll solveab(ll a, ll b) {
  ll c=0, d=0, cpd=0, lst=1;
  while(akt1>0 && Ci[akt1-1].st<=b) --akt1;
  ll po=0, ko=Ci.size()-1;
  
  /*while(po<ko) {
    ll sr=(po+ko)/2;
    if(Ci[sr].st>b) po=sr+1; else ko=sr;
  }*/
  c=Ci[akt1].nd;
  po=0; ko=Di.size()-1;
  while(po<ko) {
    ll sr=(po+ko)/2;
    if(Di[sr].st>b) po=sr+1; else ko=sr;
  }
  d=Di[po].nd;
  po=0; ko=(int)Ei.size()-1;
  while(po<ko) {
    ll sr=(po+ko)/2;
    if(Ei[sr].st<b) po=sr+1; else ko=sr;
  }
  if(Ei.size()>0 && Ei[po].st>=b) cpd=max(Ei[po].nd, 0ll);
  return a+b+max(c+d, cpd);
}
ll solvea(ll a) {
  Ci.clear(); Di.clear(); Ei.clear();
  vector<pair<ll,ll>>P;
  rep(i, n) P.pb({T[i].nd, i});
  sort(all(P));
  vector<pair<ll,ll>>Q;
  rep(i, n) {
    Q.pb({max(1ll, T[P[i].nd].st-a), T[P[i].nd].st});
    for(int j=(int)Q.size()-1; j>0; --j) if(Q[j]<Q[j-1]) swap(Q[j], Q[j-1]);
    ll ma=1, mi=0;
    for(auto j : Q) {
      mi=max(mi, j.st-ma-1);
      ma=max(ma, j.nd);
    }
    mi=max(mi, R-ma);
    if(Q[0].st!=1) mi=INF;
    Ci.pb({mi, P[i].st-1});
  }
  Q.clear();
  rep(i, n) {
    Q.pb({max(1ll, T[P[n-i-1].nd].st-a), T[P[n-i-1].nd].st});
    for(int j=(int)Q.size()-1; j>0; --j) if(Q[j]<Q[j-1]) swap(Q[j], Q[j-1]);
    ll ma=1, mi=0;
    for(auto j : Q) {
      mi=max(mi, j.st-ma-1);
      ma=max(ma, j.nd);
    }
    mi=max(mi, R-ma);
    if(Q[0].st!=1) mi=INF;
    Di.pb({mi, C-P[n-i-1].st});
  }
  ll einf=-1;
  rep(xd, 2) {
    rep(i, n) rep(j, n) {
      tmi[i][j]=R+1;
      tma[i][j]=0;
    }
    rep(i, n) {
      for(int j=i+1; j<n; ++j) {
        tmi[i][j]=tmi[i][j-1];
        if(T[P[j].nd].st>=T[P[i].nd].st) tmi[i][j]=min(tmi[i][j], T[P[j].nd].st-a);
      }
    }
    rep(i, n) {
      for(int j=i-1; j>=0; --j) {
        tma[j][i]=tma[j+1][i];
        if(T[P[j].nd].st<=T[P[i].nd].st) tma[j][i]=max(tma[j][i], T[P[j].nd].st);
      }
    }
    rep(j, n) rep(i, j) {
      if(T[P[i].nd].st<=T[P[j].nd].st) {
        if(tmi[i][j-1]<=0) continue;
        else if(tmi[i][j-1]==10*INF || tma[i+1][j]==0) einf=max(einf, abs(P[i].st-P[j].st)-1);
        //Ei.pb({10*INF, abs(P[i].st-P[j].st)-1});
        else if(tmi[i][j-1]-tma[i+1][j]-2>=0) Ei.pb({tmi[i][j-1]-tma[i+1][j]-2, abs(P[i].st-P[j].st)-1});
      }
    }
    reverse(all(P));
  }
  Ei.pb({10*INF, einf});
  sort(all(Ei));
  for(int i=(int)Ei.size()-2; i>=0; --i) Ei[i].nd=max(Ei[i].nd, Ei[i+1].nd);
  vector<ll>A;
  rep(i, n) A.pb(T[i].st);
  sort(all(A));
  ll b=R-A[n-1], apb=0;
  rep(i, n-1) apb=max(apb, A[i+1]-A[i]-1);
  ll ans=INF;
  vector<ll>kol;
  rep(i, n) {
    ll x=R-T[i].st;
    if(x>=b && a+x>=apb) kol.pb(x);
  }
  rep(i, n) rep(j, i) {
    ll x=abs(T[i].st-T[j].st)-1;
    if(x==-1 || x<apb) continue;
    ll y=x-a;
    if(y>=b) kol.pb(y);
  }
  sort(all(kol));
  akt1=Ci.size();
  akt2=Di.size();
  akt3=0;
  ll lst=-INF;
  for(auto i : kol) if(i!=lst) {
    ans=min(ans, solveab(a, i));
    lst=i;
  }
  return ans;
}
ll solve() {
  ll ans=INF;
  rep(i, n) ans=min(ans, solvea(T[i].st-1));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0); cin.tie(0);
  cin >> R >> C >> n;
  rep(i, n) cin >> T[i].st >> T[i].nd;
  ll ans=INF;
  rep(i, 4) {
    ans=min(ans, solve());
    rep(j, n) T[j]={C-T[j].nd+1, T[j].st};
    swap(R, C);
  }
  cout << ans << '\n';
}

Compilation message (stderr)

cultivation.cpp: In function 'll solveab(ll, ll)':
cultivation.cpp:16:23: warning: unused variable 'lst' [-Wunused-variable]
   16 |   ll c=0, d=0, cpd=0, lst=1;
      |                       ^~~

• Subtask 1: 5 / 5
• Subtask 2: 0 / 10
• Subtask 3: 0 / 15
• Subtask 4: 30 / 30
• Subtask 5: 20 / 20
• Subtask 6: 0 / 20