제출 #1220025

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1220025		brinton	DNA 돌연변이 (IOI21_dna)	C++20	100 / 100	25 ms	5056 KiB

dna

#include "dna.h"
#include <bits/stdc++.h>
using namespace std;

enum dnaType{
	AT,AC,TA,TC,CA,CT
};
vector<vector<int>> pref;
void init(string a, string b) {
	int N = a.size();
	pref = vector<vector<int>>(6,vector<int>(N,0));
	for(int i = 0;i < N;i++){
		if(a[i] == b[i]) continue;
		if(a[i] == 'A'){
			if(b[i] == 'T') pref[AT][i] = 1;
			else pref[AC][i] = 1;
		}else if(a[i] == 'T'){
			if(b[i] == 'A') pref[TA][i] = 1;
			else pref[TC][i] = 1;
		}else if(a[i] == 'C'){
			if(b[i] == 'A') pref[CA][i] = 1;
			else pref[CT][i] = 1;
		}
	}
	for(int i = 1;i < N;i++){
		for(int type = 0;type < 6;type++){
			pref[type][i] += pref[type][i-1];
		}
	}
}

int get_distance(int x, int y) {
	vector<int> cnt(6);
	for(int i = 0;i < 6;i++){
		if(x > 0) cnt[i] = pref[i][y]-pref[i][x-1];
		else cnt[i] = pref[i][y];
	}
	
	if(cnt[AT]+cnt[AC] != cnt[TA]+cnt[CA]) return -1;
	if(cnt[TA]+cnt[TC] != cnt[AT]+cnt[CT]) return -1;
	if(cnt[CA]+cnt[CT] != cnt[AC]+cnt[TC]) return -1;
	int step = 0;
	int m1 = min(cnt[AT],cnt[TA]);
	int m2 = min(cnt[AC],cnt[CA]);
	int m3 = min(cnt[CT],cnt[TC]);
	step += m1+m2+m3;
	cnt[AT] -= m1;
	cnt[TA] -= m1;
	cnt[AC] -= m2;
	cnt[CA] -= m2;
	cnt[CT] -= m3;
	cnt[TC] -= m3;

	assert((cnt[TA] == cnt[AC] && cnt[AC] == cnt[CT]) || 
		(cnt[AT] == cnt[CA] && cnt[CA] == cnt[TC]));
	step += (cnt[AT]+cnt[TA]+cnt[AC]+cnt[CA]);
	return step;
}
// AT,AC,TA,TC,CA,CT

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