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  1. /******************************
  2. *    author : @marvinthang    *
  3. *    date : 26 / 01 / 2022    *
  4. ******************************/
  5.  
  6. #include <bits/stdc++.h>
  7.  
  8. using namespace std;
  9.  
  10. #define  superspeed  ios_base::sync_with_stdio(false); cin.tie(nullptr); //cout.tie(nullptr);
  11. #define  file(name)  if (fopen (name".inp", "r") ) { freopen (name".inp", "r", stdin); freopen (name".out", "w", stdout); }
  12.  
  13. template <class U, class V> ostream & operator << (ostream& out, const pair<U, V> &p) { return out << '(' << p.first << ", " << p.second << ')'; }
  14. template <class T> ostream & operator << (ostream &out, const vector<T> &vt) { out << '{'; for (size_t i = 0; i + 1 < vt.size(); i++) out << vt[i] << ", "; if (!vt.empty()) out << vt.back(); return out << '}'; }
  15.  
  16. const 		int MOD = 1e9 + 7;
  17. const     double PI = 3.1415926535897932384626433832795; // acos(-1.0); atan(-1.0);
  18. const     int dir[] = {0, 1, 0, -1, 0}; // {0, 1, 1, -1, -1, 1, 0, -1, 0};
  19. const  long long oo = 1e18;
  20. const       int MAX = 1e5 + 5;
  21. const 		int LOG = 20;
  22.  
  23. int N, pos[MAX], mi[MAX][LOG], ma[MAX][LOG], lg2[MAX];
  24. int F[MAX];
  25.  
  26. int dist(int a, int b) {
  27.     int res = a - b;
  28.     if (res < 0) return -res;
  29.     return res;
  30. }
  31.  
  32. void rmq(void) {
  33. 	for (int i = 1; i <= N; ++i) {
  34. 		mi[i][0] = ma[i][0] = pos[i];
  35. 		lg2[i] = lg2[i - 1];
  36. 		while (1 << lg2[i] <= i) ++lg2[i]; --lg2[i];
  37. 	}
  38. 	for (int k = 1; k < LOG; ++k) {
  39. 		for (int i = 1; i + (1 << k) - 1 <= N; ++i) {
  40. 			mi[i][k] = min(mi[i][k - 1], mi[i + (1 << k - 1)][k - 1]);
  41. 			ma[i][k] = max(ma[i][k - 1], ma[i + (1 << k - 1)][k - 1]);
  42. 		}
  43. 	}
  44. }
  45.  
  46. int getMin(int l, int r) {
  47. 	assert(l <= r);
  48. 	int k = lg2[r - l + 1];
  49. 	return min(mi[l][k], mi[r - (1 << k) + 1][k]);
  50. }
  51.  
  52. int getMax(int l, int r) {
  53. 	assert(l <= r);
  54. 	int k = lg2[r - l + 1];
  55. 	return max(ma[l][k], ma[r - (1 << k) + 1][k]);
  56. }
  57.  
  58. bool check(int d) {
  59. 	memset(F, 0, sizeof(F));
  60. 	F[1] = 1; F[2] = -1;
  61. 	for (int i = 1; i < N; ++i) {
  62. 		F[i] += F[i - 1];
  63. 		if (!F[i]) return false;
  64. 		// int tmp = i;
  65. 		int l = i, r = N;
  66. 		while (l <= r) {
  67. 			int mid = l + r >> 1;
  68. 			if (dist(getMin(i, mid), pos[i]) <= d && dist(getMax(i, mid), pos[i]) <= d) l = mid + 1;
  69. 			else r = mid - 1;
  70. 		}
  71. 		// while (tmp <= N && dist(pos[tmp], pos[i]) <= d) ++tmp; --tmp;
  72. 		++F[i + 1]; --F[r + 1];
  73. 	}
  74. 	return F[N] + F[N - 1];
  75. }
  76.  
  77. int main(void) {
  78.     superspeed;
  79.     file("vnuoi22_post");
  80.     cin >> N; N += 2;
  81.     for (int i = 1; i <= N; ++i) cin >> pos[i];
  82.     rmq();
  83.     int l = dist(pos[1], pos[2]), r = 1e9;
  84. 	while (l <= r) {
  85. 		int mid = l + r >> 1;
  86. 		if (check(mid)) {
  87. 			r = mid - 1;
  88. 		} else l = mid + 1;
  89. 	}
  90. 	cout << l << '\n';
  91.     return 0;
  92. }
  93.  
Success #stdin #stdout 0.01s 6132KB
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https://ideone.com/dgN6zP
language:
C++14 (gcc 8.3)
created:
3 years ago
visibility:
public

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