usaco

Clean implementations of solutions to USACO problems

  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15
  16. 16
  17. 17
  18. 18
  19. 19
  20. 20
  21. 21
  22. 22
  23. 23
  24. 24
  25. 25
  26. 26
  27. 27
  28. 28
  29. 29
  30. 30
  31. 31
  32. 32
  33. 33
  34. 34
  35. 35
  36. 36
  37. 37
  38. 38
  39. 39
  40. 40
  41. 41
  42. 42
  43. 43
  44. 44
  45. 45
  46. 46
  47. 47
  48. 48
  49. 49
  50. 50
  51. 51
  52. 52
  53. 53
  54. 54
  55. 55
  56. 56
  57. 57
#include <algorithm>
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
constexpr auto INF = (int)1e9;
constexpr auto MAXN = 100005;

int N, A[100005], B[100005], R[100005], seg[4 * MAXN] = { 0 };

void update(int x, int v, int l = 0, int r = -1, int n = 1) {
	if (r == -1) r = N - 1;
	if (l == r) seg[n] = max(v, seg[n]);
	else {
		int m = (l + r) >> 1;
		x <= m ? update(x, v, l, m, n << 1) : update(x, v, m + 1, r, n << 1 | 1);
		seg[n] = max(seg[n << 1], seg[n << 1 | 1]);
	}
}

int query(int a, int b, int l = 0, int r = -1, int n = 1) {
	if (r == -1) r = N - 1;
	if (l > b || r < a) return 0;
	if (l >= a && r <= b) return seg[n];
	int m = (l + r) >> 1;
	return max(query(a, b, l, m, n << 1), query(a, b, m + 1, r, n << 1 | 1));
}

int main() {
	ifstream cin("nocross.in");
	ofstream cout("nocross.out");

	cin >> N;
	for (int i = 0; i < N; ++i) cin >> A[i], A[i]--;
	for (int i = 0; i < N; ++i) cin >> B[i], B[i]--;
	for (int i = 0; i < N; ++i) R[B[i]] = i;

	for (int i = 0; i < N; ++i) {
		vii u;
		for (int j = max(A[i] - 4, 0); j < min(A[i] + 5, N); ++j) u.emplace_back(R[j], query(0, R[j] - 1) + 1);
		for (auto q : u) update(q.first, q.second);
	}

	cout << query(0, N - 1) << endl;
}