Cod sursa(job #2786698)

Utilizator YusyBossFares Yusuf YusyBoss Data 21 octombrie 2021 15:18:20
Problema Lowest Common Ancestor Scor 0
Compilator cpp-64 Status done
Runda Arhiva educationala Marime 1.99 kb
#include <fstream>
#include <vector>
#define LOG2MAX 16
#define NMAX 100000

using namespace std;

ifstream cin ("lca.in");
ofstream cout ("lca.out");

struct strnod {
  vector <int> vfii;
};

struct streuler {
  int val, inaltime;
};

struct strrmq {
  int val, poz;
};

int neuler;
int vpoz[NMAX + 1], vlog2[2 * NMAX + 1];
strrmq vrmq[LOG2MAX + 1][2 * NMAX + 1];
strnod v[NMAX + 1];
streuler veuler[2 * NMAX + 1];

static inline int minim(int a, int b) {
  return a < b ? a : b;
}

static inline int maxim(int a, int b) {
  return a > b ? a : b;
}

void parc_euler(int index, int h) {
  int i, n;

  n = v[index].vfii.size();
  veuler[neuler].val = index;
  veuler[neuler++].inaltime = h;
  vpoz[index] = neuler - 1;
  for (i = 0; i < n; i++) {
    parc_euler(v[index].vfii[i], h + 1);
    veuler[neuler].val = index;
    veuler[neuler++].inaltime = h;
    vpoz[index] = neuler - 1;
  }
}

void prec_log2(int n) {
  int i;
  vlog2[1] = 0;
  for (i = 2; i <= n; i++)
    vlog2[i] = vlog2[i / 2] + 1;
}

strrmq cmp(strrmq A, strrmq B) {
  if (A.val < B.val)
    return A;
  return B;
}

int prec_rmq(int n) {
  int i, j;

  for (i = 0; i < n; i++) {
    vrmq[0][i].val = veuler[i].inaltime;
    vrmq[0][i].poz = i;
  }

  for (i = 1; (1 << i) <= n; i++)
    for (j = 0; j < n - (1 << i); j++)
      vrmq[i][j] = cmp(vrmq[i - 1][j], vrmq[i - 1][j + (1 << (i - 1))]);
}

int query_rmq(int st, int dr) {
  strrmq sol;
  sol = cmp(vrmq[vlog2[dr - st + 1]][st], vrmq[vlog2[dr - st + 1]][dr - (1 << vlog2[dr - st + 1]) + 1]);
  return sol.poz;
}

int main(){
  int n, q, i, node1, node2, tata;

  cin >> n >> q;
  for (i = 2; i <= n; i++) {
    cin >> tata;
    v[tata].vfii.push_back(i);
  }

  neuler = 0;
  parc_euler(1, 0);
  prec_log2(neuler);
  prec_rmq(neuler);

  while (q--) {
    cin >> node1 >> node2;
    cout << veuler[query_rmq(minim(vpoz[node1], vpoz[node2]), maxim(vpoz[node1], vpoz[node2]))].val << "\n";
  }
  return 0;
}