Yukicoder No.545 ママの大事な二人の子供
サンプルが弱いようには見えないがなぜかすり抜け$2$WAを生やした。
solution
半分全列挙。$A_i$と$- B_i$のどちらかを選んで足し合わせると言い換えて、絶対値を$0$に近付けるようにする。$O(2^\frac{N}{2})$。
implementation
#include <algorithm>
#include <cassert>
#include <climits>
#include <cstdio>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
#define repeat_from(i, m, n) for (int i = (m); (i) < int(n); ++(i))
#define whole(f, x, ...) ([&](decltype((x)) whole) { return (f)(begin(whole), end(whole), ## __VA_ARGS__); })(x)
using ll = long long;
using namespace std;
template <class T> inline void setmin(T & a, T const & b) { a = min(a, b); }
template <typename UnaryPredicate>
ll binsearch(ll l, ll r, UnaryPredicate p) { // [l, r), p is monotone
assert (l < r);
-- l;
while (r - l > 1) {
ll m = (l + r) / 2;
(p(m) ? r : l) = m;
}
return r; // = min { x in [l, r) | p(x) }, or r
}
vector<ll> calc_sums(int l, int r, vector<ll> const & a, vector<ll> const & b) {
vector<ll> cur, prv;
cur.push_back(0);
repeat_from (i, l, r) {
cur.swap(prv);
cur.clear();
for (ll x : prv) {
cur.push_back(x + a[i]);
cur.push_back(x - b[i]);
}
}
whole(sort, cur);
cur.erase(whole(unique, cur), cur.end());
return cur;
}
ll solve(int n, vector<ll> const & a, vector<ll> const & b) {
if (n == 1) {
return min(a[0], b[0]);
}
vector<ll> left = calc_sums(0, n / 2, a, b);
vector<ll> right = calc_sums(n / 2, n, a, b);
ll result = LLONG_MAX;
for (ll x : left) {
int i = binsearch(0, right.size(), [&](int i) {
return x + right[i] >= 0;
});
for (int di = -1; di <= +1; ++ di) {
if (0 <= i + di and i + di < right.size()) {
setmin(result, abs(x + right[i + di]));
}
}
}
return result;
}
int main() {
int n; scanf("%d", &n);
vector<ll> a(n), b(n); repeat (i, n) scanf("%lld%lld", &a[i], &b[i]);
ll result = solve(n, a, b);
printf("%lld\n", result);
return 0;
}