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>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long k, n, ans = 1e18, ss = 0, S = 0, S2 = 0;
cin >> k >> n;
vector<long long> a, b;
for (int i = 0; i < n; i++) {
char x, y;
int s, t;
cin >> x >> s >> y >> t;
if (x == y) {
ss += abs(s - t);
continue;
}
if (x == 'A') a.push_back(s), b.push_back(t);
else a.push_back(t), b.push_back(s);
}
if (a.size()) {
sort(a.begin(), a.end());
sort(b.begin(), b.end());
n = a.size();
vector<long long> pre1(n + 1), pre2(n + 1);
for (int i = 0; i < n; i++) pre1[i + 1] = pre1[i] + a[i], S += a[i];
for (int i = 0; i < n; i++) pre2[i + 1] = pre2[i] + b[i], S2 += b[i];
for (auto x : a) {
long long k = upper_bound(a.begin(), a.end(), x) - a.begin();
long long k2 = upper_bound(b.begin(), b.end(), x) - b.begin();
ans = min(ans, 2 * (x * (k + k2 - n) - pre1[k] - pre2[k2]));
}
for (auto x : b) {
long long k = upper_bound(a.begin(), a.end(), x) - a.begin();
long long k2 = upper_bound(b.begin(), b.end(), x) - b.begin();
ans = min(ans, 2 * (x * (k + k2 - n) - pre1[k] - pre2[k2]));
}
cout << ans + S + S2 + ss + n << '\n';
} else {
cout << ss << '\n';
}
}
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