Cod sursa(job #2208659)

Utilizator EclipseTepes Alexandru Eclipse Data 30 mai 2018 20:48:17
Problema Atac Scor 0
Compilator cpp Status done
Runda Arhiva de probleme Marime 2.12 kb
#include <iostream>
#include <fstream>
#include <vector>
#include <math.h>
#include <cmath>
#define dMAX 32000

using namespace std;

int n, m, p;
int x, y, c;

int X, Y, A, B, C, D, XP, YP;

int arr[dMAX * 2 + 2], idx;

vector<pair<int, int> > graf[dMAX + 2];
int first[dMAX * 2 + 2];
bool viz[dMAX + 2];

int sparseTable[dMAX * 2][(int)log2(2 * dMAX) + 1];

ifstream fin("atac.in");
ofstream fout("atac.out");

void DFS(int v, int cost) {
    viz[v] = true;
    arr[++idx] = cost;
    first[v] = idx;
    int newV, u, cc;
    for (u = 0; u < graf[v].size(); u++) {
        newV = graf[v][u].first;
        cc   = graf[v][u].second;
        if (!viz[newV]) {
            viz[newV] = true;
            DFS(newV, cc);
            arr[++idx] = cc;
        }
    }
}

void MakeTable() {
    int i, j, l, r;
    for (i = 1; i <= idx; i++) {
        sparseTable[i][0] = i;
    }
    for (j = 1; (1 << j) <= idx; j++) {
        for (i = 1; i + (1 << j) - 1 <= idx; i++) {
            l = sparseTable[i][j - 1];
            r = sparseTable[i + (1 << (j - 1))][j - 1];
            if (arr[l] < arr[r]) {
                   sparseTable[i][j] = l;
            } else sparseTable[i][j] = r;
        }
    }
}

int RMQ(int low, int r) {
    int l = r - low + 1;
    int k = (int)log2(l);
    int t1, t2;
    t1 = sparseTable[low][k];
    t2 = sparseTable[low + l - (1 << k)][k];
    return min(arr[t1], arr[t2]);
}

int LCE(int p, int q) {
    p = first[p];
    q = first[q];
    if (p > q) swap(p, q);
    return RMQ(p, q);
}

int main() {
    int i, j;
    fin >> n >> m >> p;
    for (i = 2; i <= n; i++) {
        fin >> x >> c;
        graf[i].push_back({x, c});
        graf[x].push_back({i, c});
    }

    DFS(1, 0);
    arr[1] = 99999999;
    MakeTable();
    fin >> X >> Y >> A >> B >> C >> D;
    XP = X, YP = Y;
    for (i = 1; i <= m; i++) {
        if (i >= p) {
            fout << LCE(X, Y) << '\n';
        } else {
            XP = (X * A + Y * B) % n + 1;
            YP = (Y * C + LCE(X, Y) * D) % n + 1;
            X = XP, Y = YP;
        }
    }
    return 0;
}