oj.uz

답안 #535015

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
535015 2022-03-09T09:55:54 Z chenwz Regions (IOI09_regions) C++11
100 / 100
 2339 ms 45032 KB 
#include <bits/stdc++.h>
using namespace std;
using VI = vector<int>;
using IP = pair<int, int>;
using LL = long long;
struct Emp { // Employee
  int tin, reg;     // tin(dfs时间戳), reg(Region Id), low()
  int low, high;    /* Lowest&Highest ID of managees */
  VI ch;   // children
};
struct Reg {
  VI ids;                /* Sorted list of (new) employee IDs */
  /* Sorted of intervals with the same nesting level. Each pair is
   * (ID, depth) where ID is the left end-point of the interval (inclusive).
   * The right end-point is implicit from the following interval.
   */
  vector<IP> ranges;
  int depth; /* Working depth during the DFS. */
};
int N, R, Q;
vector<Emp> nodes;
vector<Reg> RS;
/* Does a pre-order walk over the subtree rooted at root. id_pool contains
 * the next unused employee ID, and on return it will be updated to again
 * be the next available ID.
 * This procedure builds the RS arrays, after which the tree is no longer needed.
 */
void dfs(int u, int &timer) {
  auto& r = RS[nodes[u].reg];
  r.ids.push_back(timer++);
  /* Depth changed, so after this point we need a new range */
  r.ranges.push_back(make_pair(timer, ++r.depth));
  for (int v : nodes[u].ch)
    dfs(v, timer);
  /* Undo the depth change, and start another interval after the last managee. */
  r.ranges.push_back(make_pair(timer, --r.depth));
}
/* Query in O(R2 log R1) time, by counting for each employee in r2. */
LL query_by_id(const Reg &r1, const Reg &r2) {
  LL ans = 0;
  for (int pos : r2.ids) {
    /* Find the first range that starts at pos or later. This will
     * actually be the range after the one we want.
     */
    auto it = lower_bound(r1.ranges.begin(), r1.ranges.end(), make_pair(pos, INT_MAX));
    if (it != r1.ranges.begin())
      ans += prev(it)->second;
  }
  return ans;
}

/* Query in O(R1 log R2) time, by counting for each employee in r1 */
LL query_by_range(const Reg &r1, const Reg &r2) {
  LL ans = 0;
  for (size_t i = 0; i + 1 < r1.ranges.size(); i++) {
    int p1 = r1.ranges[i].first, p2 = r1.ranges[i + 1].first;
    /* Each employee from r2 in [p1, p2) has depth managers
     * from r1. Find the intersections of [p1, p2) with the
     * employee list for r2.
     */
    auto it1 = lower_bound(r2.ids.begin(), r2.ids.end(), p1),
         it2 = lower_bound(r2.ids.begin(), r2.ids.end(), p2);
    ans += r1.ranges[i].second * (it2 - it1);
  }
  return ans;
}

/* Query in O(R1 + R2) time, by counting for each employee in r1
 * but with a linear sweep instead of a binary search.
 */
LL query_stitch(const Reg &r1, const Reg &r2) {
  LL ans = 0;
  /* Find the first employee id that is in the first range */
  auto id = r2.ids.begin();
  if (r1.ranges.empty()) return ans;
  while (id != r2.ids.end() && *id < r1.ranges[0].first) id++;
  /* Iterate over the ranges as above */
  for (size_t i = 0; i + 1 < r1.ranges.size() && id != r2.ids.end(); i++) {
    /* Find the end of the section of employees from this range */
    auto id_bak = id;
    while (id != r2.ids.end() && *id < r1.ranges[i + 1].first)
      id++;
    ans += r1.ranges[i].second * (id - id_bak);
  }
  return ans;
}
LL solve(int r1, int r2) {
  static map<IP, LL> cache;
  IP key(r1, r2);
  if (cache.count(key)) return cache[key];
  const Reg &reg1 = RS[r1], &reg2 = RS[r2];
  int sz1 = reg1.ids.size(), sz2 = reg2.ids.size();
  int costs[3] = {
    sz1 * ((int)log2(sz2) + 2) * 5, sz2 * ((int)log2(sz1) + 2) * 5, sz1 + sz2
  };
  int k = min_element(costs, costs + 3) - costs;
  if (k == 0) return cache[key] = query_by_range(reg1, reg2);
  if (k == 1) return cache[key] = query_by_id(reg1, reg2);
  return cache[key] = query_stitch(reg1, reg2);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> N >> R >> Q, nodes.resize(N), RS.resize(R);
  cin >> nodes[0].reg, --nodes[0].reg;
  for (int i = 1, fa; i < N; i++) {
    cin >> fa >> nodes[i].reg;
    --fa, --nodes[i].reg, nodes[fa].ch.push_back(i);
  }
  int id_pool = 0;
  dfs(0, id_pool);
  for (int q = 0, r1, r2; q < Q; q++) {
    cin >> r1 >> r2, r1--, r2--;
    cout << solve(r1, r2) << endl;
  }
  return 0;
}

Subtask #1
15.0 / 15.0

# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 200 KB Output is correct
2 Correct 1 ms 200 KB Output is correct
3 Correct 3 ms 200 KB Output is correct
4 Correct 4 ms 328 KB Output is correct
5 Correct 9 ms 400 KB Output is correct
6 Correct 22 ms 752 KB Output is correct
7 Correct 15 ms 604 KB Output is correct
8 Correct 34 ms 636 KB Output is correct
9 Correct 54 ms 1344 KB Output is correct
10 Correct 66 ms 1708 KB Output is correct
11 Correct 116 ms 2352 KB Output is correct
12 Correct 182 ms 3284 KB Output is correct
13 Correct 169 ms 3372 KB Output is correct
14 Correct 181 ms 4004 KB Output is correct
15 Correct 243 ms 7840 KB Output is correct

Subtask #2
85.0 / 85.0

# 결과 실행 시간 메모리 Grader output
1 Correct 850 ms 10004 KB Output is correct
2 Correct 944 ms 8884 KB Output is correct
3 Correct 1441 ms 14960 KB Output is correct
4 Correct 241 ms 5144 KB Output is correct
5 Correct 440 ms 8044 KB Output is correct
6 Correct 585 ms 7764 KB Output is correct
7 Correct 1006 ms 9120 KB Output is correct
8 Correct 1040 ms 18956 KB Output is correct
9 Correct 1648 ms 25832 KB Output is correct
10 Correct 2171 ms 34340 KB Output is correct
11 Correct 2122 ms 31412 KB Output is correct
12 Correct 1060 ms 24868 KB Output is correct
13 Correct 1328 ms 27492 KB Output is correct
14 Correct 1822 ms 28984 KB Output is correct
15 Correct 2259 ms 37372 KB Output is correct
16 Correct 2261 ms 45032 KB Output is correct
17 Correct 2339 ms 42936 KB Output is correct