結果
問題 | No.1332 Range Nearest Query |
ユーザー |
|
提出日時 | 2021-11-10 03:31:40 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 335 ms / 2,500 ms |
コード長 | 5,106 bytes |
コンパイル時間 | 2,726 ms |
コンパイル使用メモリ | 211,656 KB |
最終ジャッジ日時 | 2025-01-25 15:15:22 |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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ファイルパターン | 結果 |
---|---|
other | AC * 48 |
ソースコード
#include <bits/stdc++.h>using namespace std;namespace atcoder {namespace internal {// @param n `0 <= n`// @return minimum non-negative `x` s.t. `n <= 2**x`int ceil_pow2(int n) {int x = 0;while ((1U << x) < (unsigned int)(n)) x++;return x;}// @param n `1 <= n`// @return minimum non-negative `x` s.t. `(n & (1 << x)) != 0`constexpr int bsf_constexpr(unsigned int n) {int x = 0;while (!(n & (1 << x))) x++;return x;}// @param n `1 <= n`// @return minimum non-negative `x` s.t. `(n & (1 << x)) != 0`int bsf(unsigned int n) {#ifdef _MSC_VERunsigned long index;_BitScanForward(&index, n);return index;#elsereturn __builtin_ctz(n);#endif}} // namespace internal} // namespace atcodernamespace atcoder {template <class S, S (*op)(S, S), S (*e)()> struct segtree {public:segtree() : segtree(0) {}explicit segtree(int n) : segtree(std::vector<S>(n, e())) {}explicit segtree(const std::vector<S>& v) : _n(int(v.size())) {log = internal::ceil_pow2(_n);size = 1 << log;d = std::vector<S>(2 * size, e());for (int i = 0; i < _n; i++) d[size + i] = v[i];for (int i = size - 1; i >= 1; i--) {update(i);}}void set(int p, S x) {assert(0 <= p && p < _n);p += size;d[p] = x;for (int i = 1; i <= log; i++) update(p >> i);}S get(int p) const {assert(0 <= p && p < _n);return d[p + size];}S prod(int l, int r) const {assert(0 <= l && l <= r && r <= _n);S sml = e(), smr = e();l += size;r += size;while (l < r) {if (l & 1) sml = op(sml, d[l++]);if (r & 1) smr = op(d[--r], smr);l >>= 1;r >>= 1;}return op(sml, smr);}S all_prod() const { return d[1]; }template <bool (*f)(S)> int max_right(int l) const {return max_right(l, [](S x) { return f(x); });}template <class F> int max_right(int l, F f) const {assert(0 <= l && l <= _n);assert(f(e()));if (l == _n) return _n;l += size;S sm = e();do {while (l % 2 == 0) l >>= 1;if (!f(op(sm, d[l]))) {while (l < size) {l = (2 * l);if (f(op(sm, d[l]))) {sm = op(sm, d[l]);l++;}}return l - size;}sm = op(sm, d[l]);l++;} while ((l & -l) != l);return _n;}template <bool (*f)(S)> int min_left(int r) const {return min_left(r, [](S x) { return f(x); });}template <class F> int min_left(int r, F f) const {assert(0 <= r && r <= _n);assert(f(e()));if (r == 0) return 0;r += size;S sm = e();do {r--;while (r > 1 && (r % 2)) r >>= 1;if (!f(op(d[r], sm))) {while (r < size) {r = (2 * r + 1);if (f(op(d[r], sm))) {sm = op(d[r], sm);r--;}}return r + 1 - size;}sm = op(d[r], sm);} while ((r & -r) != r);return 0;}private:int _n, size, log;std::vector<S> d;void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }};} // namespace atcoderint op1(int a, int b) {return min(a, b);}int e1() {return (int)(2e9);}int op2(int a, int b) {return max(a, b);}int e2() {return (int)(-2e9);}using namespace atcoder;int main(void){cin.tie(0);ios::sync_with_stdio(false);int N,Q,L,R,X;cin >> N;segtree<int, op1, e1> minseg(N+1);segtree<int, op2, e2> maxseg(N+1);vector <pair<int,pair<pair<int,int>,int>>> v;for(int i=1;i<=N;i++){cin >> X;v.push_back(make_pair(X,make_pair(make_pair(i,i),-1)));}cin >> Q;vector <int> res(Q,2e9);for(int i=0;i<Q;i++){cin >> L >> R >> X;v.push_back(make_pair(X,make_pair(make_pair(L,R),i)));}sort(v.begin(),v.end());for(int i=0;i<v.size();i++){if(v[i].second.second==-1){maxseg.set(v[i].second.first.first,v[i].first);}else{int L = v[i].second.first.first;int R = v[i].second.first.second;int idx = v[i].second.second;int val = maxseg.prod(L,R+1);if(val<=-2e9) continue;res[idx] = min(res[idx],v[i].first - val);}}sort(v.rbegin(),v.rend());for(int i=0;i<v.size();i++){if(v[i].second.second==-1){minseg.set(v[i].second.first.first,v[i].first);}else{int L = v[i].second.first.first;int R = v[i].second.first.second;int idx = v[i].second.second;int val = minseg.prod(L,R+1);if(val>=2e9) continue;res[idx] = min(res[idx],val - v[i].first);}}for(int i=0;i<Q;i++){cout << res[i] << '\n';}return 0;}