Submission #791194

#	Time	Username	Problem	Language	Result	Execution time	Memory
791194		shoryu386	Mutating DNA (IOI21_dna)	C++17	100 / 100	42 ms	8580 KiB

dna

#include <bits/stdc++.h>
using namespace std;
#define MAX 100001
int apsum[MAX][2], tpsum[MAX][2], cpsum[MAX][2];
tuple<int, int, int, int, int, int> meow[MAX];
//AT, TA, AC, CA, TC, CT
void init(string a, string b){
	int acarry1=0, tcarry1=0, ccarry1=0, acarry2=0, tcarry2=0, ccarry2=0;
	int at=0, ta=0, ac=0, ca=0, tc=0, ct=0;
	for (int x = 0; x < (int)a.size(); x++){
		if (a[x] == 'A') acarry1++;
		else if (a[x] == 'T') tcarry1++;
		else if (a[x] == 'C') ccarry1++;
		
		if (b[x] == 'A') acarry2++;
		else if (b[x] == 'T') tcarry2++;
		else if (b[x] == 'C') ccarry2++;
		
		apsum[x][0] = acarry1;
		tpsum[x][0] = tcarry1;
		cpsum[x][0] = ccarry1;
		
		apsum[x][1] = acarry2;
		tpsum[x][1] = tcarry2;
		cpsum[x][1] = ccarry2;
		
		if (a[x] == 'A' && b[x] == 'T') at++;
		else if (a[x] == 'T' && b[x] == 'A') ta++;
		else if (a[x] == 'A' && b[x] == 'C') ac++;
		else if (a[x] == 'C' && b[x] == 'A') ca++;
		else if (a[x] == 'T' && b[x] == 'C') tc++;
		else if (a[x] == 'C' && b[x] == 'T') ct++;
		
		meow[x] = make_tuple(at, ta, ac, ca, tc, ct);
	}
}

int get_distance(int x, int y){
	int at=0, ta=0, ac=0, ca=0, tc=0, ct=0;
	if (x != 0){
		if (apsum[y][0]-apsum[x-1][0] != apsum[y][1]-apsum[x-1][1] ||
		tpsum[y][0]-tpsum[x-1][0] != tpsum[y][1]-tpsum[x-1][1] ||
		cpsum[y][0]-cpsum[x-1][0] != cpsum[y][1]-cpsum[x-1][1]
		){
			return -1;
		}
		at = get<0>(meow[y])-get<0>(meow[x-1]);
		ta = get<1>(meow[y])-get<1>(meow[x-1]);
		ac = get<2>(meow[y])-get<2>(meow[x-1]);
		ca = get<3>(meow[y])-get<3>(meow[x-1]);
		tc = get<4>(meow[y])-get<4>(meow[x-1]);
		ct = get<5>(meow[y])-get<5>(meow[x-1]);
	}
	else{
		if (apsum[y][0] != apsum[y][1] ||
		tpsum[y][0] != tpsum[y][1] ||
		cpsum[y][0] != cpsum[y][1]
		){
			return -1;
		}
		
		at = get<0>(meow[y]);
		ta = get<1>(meow[y]);
		ac = get<2>(meow[y]);
		ca = get<3>(meow[y]);
		tc = get<4>(meow[y]);
		ct = get<5>(meow[y]);
	}
	
	//now it is guaranteed that we are able to somehow rearrange it
	
	//we would always prefer to fix 2-cycles before hand? with one swap
	//leaving only 3-cycles that can be fixed with 2 swaps?
	
	int ans = 0;
	int sweep = min(at, ta);
	ans += sweep; at -= sweep; ta -= sweep;
	sweep = min(ac, ca);
	ans += sweep; ac -= sweep; ca -= sweep;
	sweep = min(tc, ct);
	ans += sweep; tc -= sweep; ct -= sweep;
	
	int sum = at + ta + ac + ca + tc + ct;
	ans += (sum/3)*2;
	
	return ans;
}

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