이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <numeric>
#include <algorithm>
using namespace std;
typedef vector<vector<int>> int33;
map<char, int> get_index;
map<int, char> get_char;
int n;
vector<pair<int, int>> segtree_span;
vector<int33> segtree;
vector<char> origin;
vector<char> mutated;
int33 calculate_crossmultiply(int index) {
int33 result{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}};
result[get_index[origin[index]]][get_index[mutated[index]]] = 1;
return result;
}
int33 combine(int33 a, int33 b) {
int33 result{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}};
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
result[i][j] = a[i][j] + b[i][j];
}
}
return result;
}
void build(int x, int l, int r) {
segtree_span[x] = make_pair(l, r);
if (l == r) {
segtree[x] = calculate_crossmultiply(l);
return;
}
int mid = (l + r) / 2;
build(2 * x, l, mid);
build(2 * x + 1, mid + 1, r);
segtree[x] = combine(segtree[2 * x], segtree[2 * x + 1]);
}
int33 query(int x, int l, int r) {
if ((segtree_span[x].first > r) || (segtree_span[x].second < l)) {
int33 result{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}};
return result;
}
if ((segtree_span[x].first >= l) && (segtree_span[x].second <= r)) {
return segtree[x];
}
return combine(query(2 * x, l, r), query(2 * x + 1, l, r));
}
void init(string a, string b) {
// initialize maps between amino acids and indices
get_index['A'] = 0; get_index['C'] = 1; get_index['T'] = 2;
get_char[0] = 'A'; get_char[1] = 'C'; get_char[2] = 'T';
// take in the data
n = a.size();
origin.resize(n);
mutated.resize(n);
for (int i = 0; i < n; i++) {
origin[i] = a[i];
mutated[i] = b[i];
}
// construct the segment tree
segtree_span.resize(4 * n);
segtree.resize(4 * n);
build(1, 0, n - 1);
}
int get_distance(int x, int y) {
int33 cross = query(1, x, y);
int distance = 0;
int result;
// check if it is possible to begin with:
bool possible = true;
for (int i = 0; i < 3; i++) {
possible &= (cross[i][0] + cross[i][1] + cross[i][2] == cross[0][i] + cross[1][i] + cross[2][i]);
}
if (!possible) {
return -1;
}
// go into all cases ig:
// (a, b) - (b, a) --> (a, a) - (b, b)
for (int i = 0; i < 3; i++) {
for (int j = 0; j < i; j++) {
result = min<int>(cross[i][j], cross[j][i]);
distance += result;
cross[i][j] -= result;
cross[j][i] -= result;
}
}
// (a, b) - (b, c) - (c, a) --> (a, a) - (b, b) - (c, c) in 2 moves
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
for (int k = 0; k < 3; k++) {
if ((i != j) && (j != k) && (k != i)) {
result = min<int>(cross[i][j], min<int>(cross[j][k], cross[k][i]));
distance += 2 * result;
cross[i][j] -= result;
cross[j][k] -= result;
cross[k][i] -= result;
}
}
}
}
return distance;
}
/*
int main() {
return 0;
}
*/
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