Header Ad

HackerRank DAG Queries problem solution

In this HackerRank DAG Queries problem solution we have given a directed acyclic graph with n vertices and m edges and each vertex has an integer associated with and the initial value is zero for all vertices. we need to perform q queries on this acyclic graph.

HackerRank DAG Queries problem solution


Problem solution in Java.

import java.io.*;
import java.util.*;

public class Solution {

  static class MyBitSet {
    private final static int ADDRESS_BITS_PER_WORD = 6;
    private long[] words;
    private transient int wordsInUse = 0;

    private static int wordIndex(int bitIndex) {
      return bitIndex >> ADDRESS_BITS_PER_WORD;
    }

    public MyBitSet(int nbits) {
      words = new long[wordIndex(nbits - 1) + 1];
    }

    public void clear() {
      while (wordsInUse > 0)
        words[--wordsInUse] = 0;
    }

    public void set(int bitIndex) {
      int wordIndex = wordIndex(bitIndex);
      expandTo(wordIndex);
      words[wordIndex] |= (1L << bitIndex);
    }

    private void expandTo(int wordIndex) {
      int wordsRequired = wordIndex + 1;
      if (wordsInUse < wordsRequired) {
        wordsInUse = wordsRequired;
      }
    }


    public void or(MyBitSet set) {
      int wordsInCommon = Math.min(wordsInUse, set.wordsInUse);

      if (wordsInUse < set.wordsInUse) {
        wordsInUse = set.wordsInUse;
      }

      for (int i = 0; i < wordsInCommon; i++) {
        words[i] |= set.words[i];
      }

      if (wordsInCommon < set.wordsInUse) {
        System.arraycopy(set.words, wordsInCommon, words, wordsInCommon,
            wordsInUse - wordsInCommon);
      }
    }

    public boolean get(int bitIndex) {
      int wordIndex = wordIndex(bitIndex);
      return (wordIndex < wordsInUse) && ((words[wordIndex] & (1L << bitIndex)) != 0);
    }
  }

  static class PS {
    int opt;
    int u;
    int x;
    int i;
  }

  static class QS {
    int u;
    int i;

    public QS(int u, int i) {
      this.u = u;
      this.i = i;
    }
  }

  static Set<Integer>[] graph;
  static int[] indeg;
  static int[] topo;
  static int ttot = 1;

  static void topo_dfs(int node) {
    topo[ttot++] = node;
    for (int i = ptr[node]; i > 0; i = nxt[i]) {
      if (--indeg[succ[i]] == 0) {
        topo_dfs(succ[i]);
      }
    }
  }

  static int[] nxt;
  static int[] succ;
  static int[] ptr;
  static int index = 1;

  static void addedge(int u, int v) {
    nxt[index] = ptr[u];
    ptr[u] = index;
    succ[index++] = v;
  }


  static final int B = 316;

  static int[] solve2(int[][] queries, int n, int nQue) {
    int q = queries[0].length - 1;
    int[] ans = new int[q + 1];

    QS[] que = new QS[nQue + 1];
    PS[] perform = new PS[q + 1];
    int ptot = 0;
    int qtot = 0;

    for (int i = 1; i <= q; i++) {
      perform[ptot] = new PS();
      perform[ptot].opt = queries[0][i];
      if (perform[ptot].opt <= 2) {
        perform[ptot].u = queries[1][i];
        perform[ptot].x = queries[2][i];
        perform[ptot++].i = i;
      } else {
        que[qtot++] = new QS(queries[1][i], i);
      }
      ans[i] = Integer.MAX_VALUE;
    }

    MyBitSet[] b = new MyBitSet[n + 1];
    for (int i = n; i > 0; i--) {
      b[i] = new MyBitSet(320);
    }

    boolean[] cover = new boolean[n + 1];
    int[] minVal = new int[n + 1];

    for (int l = (ptot - 1) - (ptot - 1) % B, r = ptot - 1; l >= 0; r = l - 1, l -= B) {
      for (int i = n; i > 0; i--) {
        b[i].clear();
      }

      for (int i = l; i <= r; ++i) {
        b[perform[i].u].set(i - l);
      }

      for (int i = 1; i <= n; i++) {
        for (int j = ptr[topo[i]]; j > 0; j = nxt[j]) {
          b[succ[j]].or(b[topo[i]]);
        }
      }

      Arrays.fill(cover, false);

      for (int i = l; i <= r; i++) {
        if (perform[i].opt == 1) {
          cover[perform[i].u] = true;
        }
      }

      for (int i = 1; i <= n; i++) {
        if (cover[topo[i]]) {
          for (int j = ptr[topo[i]]; j > 0; j = nxt[j]) {
            cover[succ[j]] = true;
          }
        }
      }

      Arrays.fill(minVal, Integer.MAX_VALUE);

      for (int i = l; i <= r; i++) {
        if (perform[i].opt == 2) {
          minVal[perform[i].u] = Math.min(minVal[perform[i].u], perform[i].x);
        }
      }

      for (int i = 1; i <= n; ++i) {
        for (int j = ptr[topo[i]]; j > 0; j = nxt[j]) {
          minVal[succ[j]] = Math.min(minVal[succ[j]], minVal[topo[i]]);
        }
      }


      int i = qtot;
      while (i > 0 && que[i - 1].i > perform[l].i) {
        i--;
      }
      while (i < qtot) {
        if (que[i].i < perform[r].i) {
          int j = r;
          while (perform[j].i > que[i].i) {
            --j;
          }
          for (; j >= l; j--)
            if (b[que[i].u].get(j - l)) {
              ans[que[i].i] = Math.min(ans[que[i].i], perform[j].x);
              if (perform[j].opt == 1) {
                --qtot;
                QS temp = que[i];
                que[i] = que[qtot];
                que[qtot] = temp;
                break;
              }
            }
          i += j < l ? 1 : 0;
        } else if (cover[que[i].u]) {
          int j = r;
          for (; perform[j].opt == 2 || !b[que[i].u].get(j - l); j--) {
            if (perform[j].opt == 2 && b[que[i].u].get(j - l)) {
              ans[que[i].i] = Math.min(ans[que[i].i], perform[j].x);
            }
          }
          ans[que[i].i] = Math.min(ans[que[i].i], perform[j].x);

          --qtot;
          QS temp = que[i];
          que[i] = que[qtot];
          que[qtot] = temp;

        } else {
          ans[que[i].i] = Math.min(ans[que[i].i], minVal[que[i].u]);
          i++;
        }

      }
    }
    while (qtot-- > 0) {
      ans[que[qtot].i] = 0;
    }
    return ans;
  }

  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    BufferedWriter bw = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

    StringTokenizer st = new StringTokenizer(br.readLine());

    int n = Integer.parseInt(st.nextToken());
    int m = Integer.parseInt(st.nextToken());
    int q = Integer.parseInt(st.nextToken());

    graph = new Set[n + 1];
    Set<Integer>[] parent = new Set[n + 1];

    for (int i = 1; i <= n; i++) {
      graph[i] = new HashSet<>();
      parent[i] = new HashSet<>();
    }

    for (int i = 0; i < m; i++) {
      st = new StringTokenizer(br.readLine());
      int u = Integer.parseInt(st.nextToken());
      int v = Integer.parseInt(st.nextToken());
      graph[u].add(v);
      parent[v].add(u);
    }

    int[][] queries = new int[3][q + 1];
    int[] convert = new int[n + 1];
    int nodes = 0;
    int op3 = 0;

    for (int i = 1; i <= q; i++) {
      st = new StringTokenizer(br.readLine());
      queries[0][i] = Integer.parseInt(st.nextToken());
      int u = Integer.parseInt(st.nextToken());

      if (convert[u] == 0) {
        nodes++;
        convert[u] = nodes;
      }

      queries[1][i] = convert[u];
      if (queries[0][i] <= 2) {
        int x = Integer.parseInt(st.nextToken());
        queries[2][i] = x;
      } else {
        op3++;
      }
    }

    for (int u = 1; u <= n; u++) {
      if (convert[u] == 0) {
        for (int v : parent[u]) {
          graph[v].remove(u);
          graph[v].addAll(graph[u]);
        }
        for (int v : graph[u]) {
          parent[v].remove(u);
          parent[v].addAll(parent[u]);
        }
        
        parent[u] = null;
        graph[u] = null;
      }
    }

    indeg = new int[nodes + 1];
    boolean[] existDeg = new boolean[nodes + 1];

    nxt = new int[m + 1];
    ptr = new int[nodes + 1];
    succ = new int[m + 1];

    for (int u1 = 1; u1 <= n; u1++) {
      int u = convert[u1];
      if (u > 0) {
        for (int v1 : graph[u1]) {
          int v = convert[v1];
          indeg[v]++;
          existDeg[v] = true;
          addedge(u, v);
        }
      }
    }

    topo = new int[nodes + 1];
    for (int i = nodes; i > 0; i--) {
      if (!existDeg[i]) {
        topo_dfs(i);
      }
    }

    int[] ans = solve2(queries, nodes, op3);

    for (int i = 1; i <= q; i++) {
      if (ans[i] < Integer.MAX_VALUE) {
        bw.write(ans[i] + "\n");
      }
    }

    bw.close();
    br.close();
  }
}

{"mode":"full","isActive":false}


Problem solution in C++.

#include <bits/stdc++.h>
using namespace std;


typedef long long ll;

#define tm suka

const int maxc = 1e9;

const int maxn = 100100;
vector<int> g[maxn];

int tm[maxn];

int qt[maxn];
int qu[maxn];
int qx[maxn];
int qans[maxn];

const int sn = 600;
const int sn2 = 1000;
const int block = maxn / 3 + 2;
bitset<block> bs[maxn];
bool used[maxn];
int L, R;
int n;

vector<int> ord;

void pre(int u) {
    used[u] = true;
    for (int v: g[u])
        if (!used[v])
            pre(v);
    ord.push_back(u);
}

void uin(int &a, int b) {
    a = min(a, b);
}

void uax(int &a, int b) {
    a = max(a, b);
}

int a[maxn / sn + 2][maxn];

int upd[maxn];
vector<pair<int, int>> o;
void put1(int u, int x) {
    o.emplace_back(u, x);
    if ((int)o.size() < sn2)
        return;
    for (int i = 0; i < n; ++i)
        upd[i] = -1;
    for (auto p: o)
        uax(upd[p.first], p.second);
    o.clear();
    for (int ii = n - 1; ii >= 0; --ii) {
        int u = ord[ii];
        tm[u] = max(tm[u], upd[u]);
        for (int v: g[u])
            uax(upd[v], upd[u]);
    }
}

int get1(int u) {
    int res = tm[u];
    for (auto op: o)
        if (bs[op.first][u - L])
            uax(res, op.second);
    return res;
}

vector<int> q2;

void calc2(int l, int r) {
    int k = l / sn;
    for (int i = 0; i < n; ++i)
        a[k][i] = maxc;
    for (int i = l; i < r; ++i) {
        int id = q2[i];
        uin(a[k][qu[id]], qx[id]);
    }
    for (int ii = n - 1; ii >= 0; --ii) {
        int u = ord[ii];
        for (int v: g[u])
            uin(a[k][v], a[k][u]);
    }
}

int get2(int u, int l, int r) {
    l = lower_bound(q2.begin(), q2.end(), l) - q2.begin();
    r = lower_bound(q2.begin(), q2.end(), r) - q2.begin();
    int res = maxc;
    int l2 = ((l + sn - 1) / sn) * sn;
    int r2 = (r / sn) * sn;
    if (l2 > r2) {
        for (int i = l; i < r; ++i) {
            int id = q2[i];
            int v = qu[id];
            if (bs[v][u - L])
                uin(res, qx[id]);
        }
    } else {
        for (int i = l; i < l2; ++i) {
            int id = q2[i];
            int v = qu[id];
            if (bs[v][u - L])
                uin(res, qx[id]);
        }
        for (int i = r2; i < r; ++i) {
            int id = q2[i];
            int v = qu[id];
            if (bs[v][u - L])
                uin(res, qx[id]);
        }
        l2 /= sn, r2 /= sn;
        for (int i = l2; i < r2; ++i)
            uin(res, a[i][u]);
    }
    return res;
}

int main() {
    int m, q;
    cin >> n >> m >> q;

    for (int i = 0; i < m; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        --u, --v;
        g[u].push_back(v);
    }
    for (int i = 0; i < n; ++i)
        if (!used[i])
            pre(i);

    for (int i = 0; i < q; ++i) {
        int t, u, x = -1;
        scanf("%d%d", &t, &u);
        --u;
        if (t != 3)
            scanf("%d", &x);
        qt[i] = t, qu[i] = u, qx[i] = x;
    }

    for (int i = 0; i < q; ++i) {
        if (qt[i] != 2)
            continue;
        q2.push_back(i);
    }
    for (int i = 0; i + sn <= (int)q2.size(); i += sn)
        calc2(i, i + sn);

    for (L = 0; L < n; L += block) {
        R = min(n, L + block);
        for (int i = 0; i < n; ++i) {
            bs[i].reset();
            tm[i] = -1;
        }
        o.clear();

        for (int i = 0; i < n; ++i) {
            int u = ord[i];
            if (L <= u && u < R)
                bs[u][u - L] = true;
            for (int v: g[u])
                bs[u] |= bs[v];
        }

        for (int id = 0; id < q; ++id) {
            if (qt[id] == 3) {
                int u = qu[id];
                if (u < L || u >= R)
                    continue;
                int lb = get1(u);
                int val = lb >= 0 ? qx[lb] : 0;
                uin(val, get2(u, lb, id));
                qans[id] = val;
            }
            if (qt[id] == 1)
                put1(qu[id], id);
        }
    }
    for (int i = 0; i < q; ++i) {
        if (qt[i] == 3)
            cout << qans[i] << '\n';
    }
}

{"mode":"full","isActive":false}


Post a Comment

0 Comments