Submission #885982

#	Time	Username	Problem	Language	Result	Execution time	Memory
885982		epicci23	Mutating DNA (IOI21_dna)	C++17	100 / 100	33 ms	9752 KiB

dna

#include "dna.h"
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())

/*
 2. asama icin notlar:
   1 - sinav baslayinca her problemi oku
   en kolay gozukeni cozmeye basla

   2 - bir soruya takilma
   wa aliyorsan ve sorunu bulamadiysan hemen diger probleme gec

   3 - her problemle ugras

   4 - cozebildigin subtasklari impl et ve puanini arttir

   5 - eger tum problemlerle ugrastiysan ve yapacagin bir sey kalmadi ise
   random observationlar dene veya brute force atip pattern bulmaya calis
*/


vector<array<array<int,3>,3>> pre;
vector<array<int,3>> cnt1,cnt2;

int f(char x){
  if(x=='A') return 0;
  if(x=='T') return 1;
  return 2;
}

void init(string a,string b){
  array<array<int,3>,3> cur;
  array<int,3> cr1,cr2;

  for(int i=0;i<3;i++)
  	for(int j=0;j<3;j++)
  		cr1[i]=cr2[i]=cur[i][j]=0;

  int n = sz(a);
  pre.resize(n);
  cnt1.resize(n);
  cnt2.resize(n);

  for(int i=0;i<n;i++){
    array<int,2> val={f(a[i]),f(b[i])};
    cur[val[0]][val[1]]++;
    
    cr1[f(a[i])]++;
    cr2[f(b[i])]++;

    pre[i]=cur;
    cnt1[i]=cr1;
    cnt2[i]=cr2;
  }
}

int get_distance(int x,int y){

   array<array<int,3>,3> res=pre[y];
   array<int,3> cnt_a=cnt1[y];
   array<int,3> cnt_b=cnt2[y];

   if(x>0){
     for(int i=0;i<3;i++)
     	for(int j=0;j<3;j++)
     	  res[i][j]-=pre[x-1][i][j];

     for(int i=0;i<3;i++){
       cnt_a[i]-=cnt1[x-1][i];
       cnt_b[i]-=cnt2[x-1][i];
     }
   }

   if(cnt_a!=cnt_b) return -1;

   int cur=min(res[0][2],res[2][0]);
   int ans=cur;

   res[0][2]-=cur;
   res[2][0]-=cur;
   
   cur=min(res[0][1],res[1][0]);
   ans+=cur;

   res[0][1]-=cur;
   res[1][0]-=cur;

   cur=min(res[1][2],res[2][1]);
   ans+=cur;

   res[1][2]-=cur;
   res[2][1]-=cur;

   int c1=min({res[0][1],res[1][2],res[2][0]});
   int c2=min({res[0][2],res[2][1],res[1][0]});

   if(c1>0 && c2>0) return -1;

   ans+=c1*2;
   ans+=c2*2;

   return ans;
}

• Subtask 1: 21 / 21
• Subtask 2: 22 / 22
• Subtask 3: 13 / 13
• Subtask 4: 28 / 28
• Subtask 5: 16 / 16