Submission #637728

#	Time	Username	Problem	Language	Result	Execution time	Memory
637728		ggoh	Aliens (IOI16_aliens)	C++14	0 / 100	1 ms	340 KiB

aliens

#include "aliens.h"
#include<bits/stdc++.h>
using namespace std;
#define sz(v) ((int)(v).size())
typedef long long lint;
typedef pair<int,int> pii;
 
vector<pii>Vtmp,V;
lint sq(int L)
{
  return 1ll*L*L;
}
int a[500004],b[500004];
struct Line{
  lint a,b;
};
vector<Line>CHT[4004];
int piv[500004];
int N;
lint D[500004],E[500004];
bool check1(Line A,Line B,Line C)
{
  if((__int128)(B.b-A.b)*(B.a-C.a) >= (__int128)(C.b-B.b)*(A.a-B.a))return 1;
  else return 0;
}
bool check2(Line A,Line B,lint x)
{
  if((__int128)(B.b-A.b) < (__int128) x*(A.a-B.a)) return 1;
  else return 0;
}
 
void insert_line(int num,Line L)
{
  while((int)CHT[num].size()>2 && check1(CHT[num][(int)CHT[num].size()-2],CHT[num][(int)CHT[num].size()-1],L)){
    CHT[num].pop_back();
  }
  CHT[num].push_back(L);
  piv[num]=min(piv[num],(int)CHT[num].size()-1);
}
lint gets(int num,lint x)
{
  while(piv[num]+1<(int)CHT[num].size() && check2(CHT[num][piv[num]],CHT[num][piv[num]+1],x))piv[num]++;
  return CHT[num][piv[num]].a * x + CHT[num][piv[num]].b;
}
lint take_photos(int n, int m, int K, std::vector<int> r, std::vector<int> c) {
  for(int i=0;i<n;i++)
  {
    if(r[i]<c[i])swap(r[i],c[i]);
    Vtmp.push_back({r[i],-c[i]});
  }
  sort(Vtmp.begin(),Vtmp.end());
  int mini=1e9;
  for(int i=(int)Vtmp.size()-1;i>=0;i--)
  {
    if(mini>-Vtmp[i].second)V.push_back({Vtmp[i].first,-Vtmp[i].second});
    mini=min(mini,-Vtmp[i].second);
  }
  sort(V.begin(),V.end());
  n=V.size();
  for(int i=1;i<=n;i++)
  {
    a[i]=V[i-1].second;
    b[i]=V[i-1].first+1;
  }
  insert_line(0,{-2*a[1],sq(a[1])});
  K=min(K,n);
  N=n;
  for(int i=0;i<=K;i++)piv[i]=0;
  for(int i=1;i<=n;i++)
  {
for(int j=1;j<=K;j++)E[j]=D[j];
    for(int k=1;k<=K;k++)
    {
      if(k>i)continue;
      D[k]=gets(k-1,b[i])+sq(b[i]);
      if(i!=n)insert_line(k,{-2*a[i+1],E[k]+sq(a[i+1])-sq(b[i]-min(b[i],a[i+1]))});
    }
  }
  lint ans=1e18;
  for(int k=1;k<=K;k++)ans=min(ans,D[k]);
  return ans;
}

• Subtask 1: 0 / 4
• Subtask 2: 0 / 12
• Subtask 3: 0 / 9
• Subtask 4: 0 / 16
• Subtask 5: 0 / 19
• Subtask 6: 0 / 40