Cod sursa(job #2666338)

Utilizator TheGodFather2131Alexandru Miclea TheGodFather2131 Data 1 noiembrie 2020 15:53:49
Problema Lowest Common Ancestor Scor 70
Compilator cpp-64 Status done
Runda Arhiva educationala Marime 1.98 kb
//ALEXANDRU MICLEA

#include <vector>
#include <algorithm>
#include <string>
#include <queue>
#include <map>
#include <set>
#include <unordered_map>
#include <time.h>
#include <iomanip>
#include <deque>
#include <math.h>
#include <cmath>
#include <assert.h>
#include <stack>
#include <bitset>
#include <random>
#include <chrono>
#include <assert.h>

using namespace std;

#include <fstream>
//ifstream cin("input.in"); ofstream cout("output.out");
ifstream cin("lca.in"); ofstream cout("lca.out");

//VARIABLES

int rmq[25][400005];
int LOG[400005];

int h[100005];
vector <vector <int> > gr(100005);
vector <int> euler;
vector <int> pos(100005);

//FUNCTIONS

void dfs(int nod) {

    euler.push_back(nod);
    pos[nod] = euler.size() - 1;

    for (auto& x : gr[nod]) {
        h[x] = h[nod] + 1;
        dfs(x);
        euler.push_back(nod);
    }
}

//MAIN
int main() {

    int n, m;
    cin >> n >> m;

    for (int i = 2; i <= 4 * n; i++) LOG[i] = LOG[i / 2] + 1;

    for (int i = 2; i <= n; i++) {
        int dad; cin >> dad;
        gr[dad].push_back(i);
    }

    dfs(1);

    for (int i = 0; i < euler.size(); i++) {
        rmq[0][i] = euler[i];
    }

    for (int p = 1; p <= LOG[4 * n]; p++) { //TODO MODIFICA DACA IA PREA MULT
        for (int i = 0; i < euler.size(); i++) {
            if (h[rmq[p - 1][i]] < h[rmq[p - 1][i + (1 << (p - 1))]]) {
                rmq[p][i] = rmq[p - 1][i];
            }
            else {
                rmq[p][i] = rmq[p - 1][i + (1 << (p - 1))];
            }
        }
    }
    
    for (int i = 1; i <= m; i++) {
        int x, y;

        cin >> x >> y;

        x = pos[x];
        y = pos[y];

        if (y < x) swap(x, y);

        int dif = y - x + 1;
        int ld = LOG[dif];

        if (h[rmq[ld][x]] < h[rmq[ld][y - (1 << ld) + 1]]) cout << rmq[ld][x];
        else cout << rmq[ld][y - (1 << ld) + 1];
        cout << '\n';

    }
    
    return 0;
}