1647 - Range Minimum Queries I
28 Jan 2021 — Tags: None
Click to show code.
using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
ll INF = 1e17;
struct segment_tree
{
int n;
vector<ll> t;
segment_tree(vi const &a)
{
n = (int)a.size();
t.resize(4 * (int)a.size());
build(a, 1, 0, n - 1);
}
ll merge(ll a, ll b) { return min(a, b); }
void build(vi const &a, int v, int tl, int tr)
{
if (tl == tr)
t[v] = a[tl];
else
{
int tm = (tl + tr) / 2;
build(a, 2 * v, tl, tm);
build(a, 2 * v + 1, tm + 1, tr);
t[v] = merge(t[2 * v], t[2 * v + 1]);
}
}
auto query(int v, int tl, int tr, int ql, int qr)
{
if (tl == ql and tr == qr)
return t[v];
else if (ql > qr)
return INF;
else
{
int tm = (tl + tr) / 2;
return merge(query(2 * v, tl, tm, ql, min(tm, qr)),
query(2 * v + 1, tm + 1, tr, max(tm + 1, ql), qr));
}
}
};
int main(void)
{
int n, q;
cin >> n >> q;
vi a(n);
for (auto &ai : a)
cin >> ai;
segment_tree st(a);
while (q--)
{
int ql, qr;
cin >> ql >> qr, ql--, qr--;
cout << st.query(1, 0, n - 1, ql, qr) << endl;
}
return 0;
}