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 ll = long long;
using std::vector;
using std::array;
using std::tuple;
using std::pair;
template <class F> struct fixed : private F {
explicit fixed(F&& f) : F(std::forward<F>(f)) {}
template <class... Args> decltype(auto) operator()(Args&&... args) const {
return F::operator()(*this, std::forward<Args>(args)...);
}
};
template <class T>
constexpr T infty = std::numeric_limits<T>::max() / 2;
constexpr pair<int, int> unit = {infty<int>, infty<int>};
class segtree {
int size;
vector<pair<int, int>> data;
public:
explicit segtree(const int n) : size(n), data(2 * n, unit) {}
void assign(int i, const pair<int, int>& x) {
i += size;
data[i] = x;
while (i > 1) {
i >>= 1;
data[i] = std::min(data[2 * i], data[2 * i + 1]);
}
}
pair<int, int> fold(int l, int r) {
l += size;
r += size;
auto ret = unit;
while (l < r) {
if (l & 1) ret = std::min(ret, data[l++]);
if (r & 1) ret = std::min(ret, data[--r]);
l >>= 1;
r >>= 1;
}
return ret;
}
};
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(nullptr);
int N, Q;
std::cin >> N >> Q;
vector<int> S(N), E(N);
for (int i = 0; i < N; ++i) {
std::cin >> S[i] >> E[i];
}
vector<pair<int, int>> vec(N);
for (int i = 0; i < N; ++i) {
vec[i] = {E[i], i};
}
std::sort(vec.begin(), vec.end());
segtree seg(N);
for (int i = 0; i < N; ++i) {
const int k = vec[i].second;
seg.assign(i, pair(S[k], k));
}
constexpr int LOG = 18;
array<vector<int>, LOG> parent = {};
parent[0].resize(N);
for (int i = 0; i < N; ++i) {
const int l = std::lower_bound(vec.begin(), vec.end(), pair(S[i], 0)) - vec.begin();
const int r = std::lower_bound(vec.begin(), vec.end(), pair(E[i] + 1, 0)) - vec.begin();
parent[0][i] = seg.fold(l, r).second;
}
for (int i = 0; i < LOG - 1; ++i) {
parent[i + 1].resize(N);
for (int j = 0; j < N; ++j) {
parent[i + 1][j] = parent[i][parent[i][j]];
}
}
while (Q--) {
int s, e;
std::cin >> s >> e;
s -= 1, e -= 1;
const int root = parent[LOG - 1][e];
if (s == e) {
std::cout << 0 << '\n';
continue;
}
if (E[e] < E[s] or E[s] < S[root]) {
std::cout << "impossible\n";
continue;
}
if (S[e] <= E[s]) {
std::cout << 1 << '\n';
continue;
}
int ans = 0;
for (int k = LOG - 1; k >= 0; --k) {
if (E[s] < S[parent[k][e]]) {
e = parent[k][e];
ans |= 1 << k;
}
}
std::cout << ans + 2 << '\n';
}
return 0;
}
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