이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "dna.h"
#include <string>
#include <vector>
#include <iostream>
using namespace std;
vector<vector<vector<int>>> pre;
int num(char c){
    if(c=='A')return 0;
    if(c=='C')return 1;
    return 2;
}
void init(string a, string b) {
    int n=a.size();
    pre=vector<vector<vector<int>>> (n+1,vector<vector<int>>(3,vector<int>(3)));
    for(int i=0;i<n;i++){
        pre[i+1]=pre[i];
        int x,y;
        x=num(a[i]);
        y=num(b[i]);
        pre[i+1][x][y]++;
    }
}
vector<vector<int>> sub(vector<vector<int>>& a,vector<vector<int>>& b){
    vector<vector<int>> ans(3,vector<int>(3));
    for(int i=0;i<3;i++){
        for(int j=0;j<3;j++){
            ans[i][j]=b[i][j]-a[i][j];
        }
    }
    return ans;
}
int get_distance(int x, int y) {
    vector<vector<int>> dif=sub(pre[x],pre[y+1]);
    int ans=0;
    for(int i=0;i<3;i++){
        for(int j=i+1;j<3;j++){
            if(dif[i][j]){
                int d=min(dif[i][j],dif[j][i]);
                dif[i][j]-=d;
                dif[j][i]-=d;
                ans+=d;
            }
        }
    }
    //cambios en los que los dos se benefician
    //0 1 2
    //1 2 0
    //2 1 0
    int d1=min(dif[0][1],min(dif[1][2],dif[2][0])),d2=min(dif[0][2],min(dif[1][0],dif[2][1]));
    ans+=2*d1;
    ans+=2*d2;
    dif[0][1]-=d1;
    dif[1][2]-=d1;
    dif[2][0]-=d1;
    dif[0][2]-=d2;
    dif[1][0]-=d2;
    dif[2][1]-=d2;
    for(int i=0;i<3;i++){
        for(int j=0;j<3;j++){
            if(dif[i][j]&&i!=j)return -1;
        }
    }
	return ans;
}
/*int main(){
    string a,b;
    int x,y;
    cin>>a>>b;
    init(a,b);
    while(cin>>x>>y){
        cout<<get_distance(x,y)<<'\n';
    }
}*/
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