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;
class RangeMin {
public:
int size_ = 1;
vector<pair<int, int>> dat;
void init(int sz) {
while (size_ <= sz) size_ *= 2;
dat.resize(size_ * 2, make_pair(0, 0));
}
void update(int pos, pair<int, int> val) {
pos += size_;
dat[pos] = val;
while (pos >= 2) {
pos >>= 1;
dat[pos] = min(dat[pos * 2 + 0], dat[pos * 2 + 1]);
}
}
pair<int, int> query_(int l, int r, int a, int b, int u) {
if (l <= a && b <= r) return dat[u];
if (r <= a || b <= l) return make_pair(2000000000, -1);
pair<int, int> v1 = query_(l, r, a, (a + b) / 2, u * 2 + 0);
pair<int, int> v2 = query_(l, r, (a + b) / 2, b, u * 2 + 1);
return min(v1, v2);
}
pair<int, int> query(int l, int r) {
return query_(l, r, 0, size_, 1);
}
};
int N, S[1 << 19], T[1 << 19];
int Q, X[1 << 19], Y[1 << 19];
int Prev[1 << 19][22];
int solve(int stt, int goa) {
if (T[stt] > T[goa]) return -1;
if (stt == goa) return 0;
if (T[stt] >= S[goa]) return 1;
// Doubling
int cx = goa, cur = 0;
for (int d = 21; d >= 0; d--) {
int nex = Prev[cx][d];
if (S[nex] > T[stt]) { cx = nex; cur += (1 << d); }
}
if (cur >= (1 << 21)) return -1;
return cur + 2;
}
int main() {
// Step 1. Input
cin >> N >> Q;
for (int i = 1; i <= N; i++) cin >> S[i] >> T[i];
for (int i = 1; i <= Q; i++) cin >> X[i] >> Y[i];
// Step 2. Sorting
vector<tuple<int, int, int>> List;
for (int i = 1; i <= N; i++) List.push_back(make_tuple(T[i], S[i], i));
sort(List.begin(), List.end());
RangeMin Z; Z.init(List.size() + 2);
for (int i = 0; i < (int)List.size(); i++) {
Z.update(i, make_pair(get<1>(List[i]), get<2>(List[i])));
}
// Step 3. Get Prev
for (int i = 0; i < (int)List.size(); i++) {
int idx = get<2>(List[i]);
int pos1 = lower_bound(List.begin(), List.end(), make_tuple(S[idx], -1, -1)) - List.begin();
pair<int, int> ret = Z.query(pos1, i);
if (ret.first == 2000000000) Prev[idx][0] = idx;
else Prev[idx][0] = ret.second;
}
// Step 4. Doubling
for (int d = 1; d <= 21; d++) {
for (int i = 1; i <= N; i++) Prev[i][d] = Prev[Prev[i][d - 1]][d - 1];
}
// Step 5. Answer Query
for (int i = 1; i <= Q; i++) {
int ret = solve(X[i], Y[i]);
if (ret == -1) cout << "impossible" << endl;
else cout << ret << endl;
}
return 0;
}
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |