이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#include "dna.h"
using namespace std;
using ll = long long;
using vb = vector<bool>;
using vvb = vector<vb>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vc = vector<char>;
using vvc = vector<vc>;
using vs = vector<string>;
const ll mod = 1e9 + 7,inf = 1e18;
#define pb push_back
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
string a,b;
void init(string A, string B) {
a = A;
b = B;
}
struct BIT
{
int n;
vl bit;
BIT(int _n)
{
n = _n;
bit.assign(n + 1,0);
}
void update(int i,ll inc)
{
for (;i>0;i-=i&(-i))bit[i] += inc;
}
ll sum(int i)
{
ll ret = 0;
for (;i<=n;i+=i&(-i))ret += bit[i];
return ret;
}
};
ll cnt_Inversions(vi&p)
{
int n = (int)p.size();
BIT ds(n);
ll ret = 0;
for (int i = 1;i<=n;i++){
int val = p[i - 1];
ret += ds.sum(val + 1);
ds.update(val + 1,1);
}
return ret;
}
ll get_ans(string s1,string s2)
{
int n = (int)s1.size();
vvi pos(26);
vvi rpos(26);
for (int i = 0;i<n;i++)
pos[s1[i] - 'A'].pb(i);
for (int i = 0;i<n;i++)
rpos[s2[i] - 'A'].pb(i);
vi ppos(26);
vi p(n);
for (int i = 0;i<n;i++){
char cur = s1[i];
p[i] = rpos[cur - 'A'][ppos[cur - 'A']];
ppos[cur - 'A']++;
}
return cnt_Inversions(p);
}
int get_distance(int x, int y) {
map<char,vi>pos1,pos2;
for (int i = x;i<=y;i++){
pos1[a[i]].pb(i);
pos2[b[i]].pb(i);
}
if (pos1['A'].size() != pos2['A'].size() or pos1['T'].size() != pos2['T'].size()
or pos1['C'].size() != pos2['C'].size())return -1;
return get_ans(a.substr(x,y - x + 1),b.substr(x,y - x + 1));
}
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