oj.uz

Submission #980087

# Submission time Handle Problem Language Result Execution time Memory
980087 2024-05-11T22:52:01 Z asdfgrace Event Hopping (BOI22_events) C++17
25 / 100
 1032 ms 15184 KB 
#include <bits/stdc++.h>
using namespace std;

#define dbg(x) //x
#define prt(x) dbg(cerr << x)
#define pv(x) dbg(cerr << #x << " = " << x << '\n')
#define pv2(x) dbg(cerr << #x << " = " << x.first << ',' << x.second << '\n')
#define parr(x) dbg(prt(#x << " = { "); for (auto y : x) prt(y << ' '); prt("}\n");)
#define parr2(x) dbg(prt(#x << " = { "); for (auto [y, z] : x) prt(y << ',' << z << "  "); prt("}\n");)
#define parr2d(x) dbg(prt(#x << ":\n"); for (auto arr : x) {parr(arr);} prt('\n'));
#define parr2d2(x) dbg(prt(#x << ":\n"); for (auto arr : x) {parr2(arr);} prt('\n'));

const int inf = 2e9 + 5;

/*
o(n^2):
for (each event)
add edge from this one to every range that contains it
and basically
you want to find the shortest path from i to j
note - if e[i] = e[j], there will be cycles in the graph
but these aren't that useful unless you want to jump directly from i to j
so we will be dealing with a dag here
so you wouldn't want to switch from anything with the same end
so e[i] must strictly increase
note that if no contained pairs
and you do have a CONTIGIOUS range of indices that you can hop to next
how to solve in n^2?
precomputing answers?
just get all the edges
maybe run n dijkstra? is this different bc dag?
for (cur node)
for (next node)
if (reachable)
for (each node)
dist[node][next] = min(dist[node][next], dist[node][cur] + 1)
or maybe just run a dijkstra if you have all the edges and n <= 1000 and q <= 100
greedy?
if you CANNOT reach the dest from here, you should go to the latest ending event
so that it starts before the dest ends
what about when does it end
if it ends before the dest ends that's fine
it has to end after the dest starts
it cannot end after the dest ends
NOTE: while we must have s[j] <= e[i] we don't have to maximize or minimize it
then we want to pick the maximum possible e[j] <= e[d]
so we can make like a pref max for end values based on start values have to be less
than or equal to whatever
but that won't factor in the "e[j] <= e[d]"
but that's a good reformulation
given max start, what is max e[j] <= e[d]?
we should be able to do queries offline then
so that for each query
we make some new values valid
*/

struct SegTree {
  int n;
  vector<int> st;
  
  SegTree (int x) {
    n = x;
    st.resize(2 * n, 0);
  }
  
  void upd(int k, int x) {
    k += n;
    st[k] = x;
    for (k /= 2; k >= 1; k /= 2) {
      st[k] = max(st[k * 2], st[k * 2 + 1]);
    }
  }
  
  int quer(int l, int r) {
    l += n; r += n;
    int res = 0;
    while (r >= l && l >= 1) {
      if (l % 2 == 1) res = max(res, st[l++]);
      if (r % 2 == 0) res = max(res, st[r--]);
      l /= 2; r /= 2;
    }
    return res;
  }
};

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<array<int, 3>> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i][0] >> a[i][1];
    a[i][2] = i;
  }
  sort(a.begin(), a.end(), [&] (array<int, 3> x, array<int, 3> y) {
    return x[1] < y[1] || (x[1] == y[1] && x[0] < y[0]);
  });
  vector<int> at(n);
  for (int i = 0; i < n; i++) {
    at[a[i][2]] = i;
  }
  if (q <= 100) {
    vector<array<int, 3>> quer(q);
    for (int i = 0; i < q; i++) {
      cin >> quer[i][0] >> quer[i][1];
      quer[i][0]--; quer[i][1]--;
      quer[i][0] = at[quer[i][0]];
      quer[i][1] = at[quer[i][1]];
      quer[i][2] = i;
    }
    sort(quer.begin(), quer.end(), [&] (array<int, 3> x, array<int, 3> y) {
      return x[1] < y[1];
    });
    parr2d(quer);
    vector<int> ans(q, -1);
    SegTree pmx(n);
    vector<int> s(n);
    for (int i = 0; i < n; i++) {
      s[i] = a[i][0];
    }
    sort(s.begin(), s.end());
    vector<int> ords(n);
    iota(ords.begin(), ords.end(), 0);
    sort(ords.begin(), ords.end(), [&] (int x, int y) {
      return a[x][0] < a[y][0];
    });
    vector<int> ats(n);
    for (int i = 0 ; i < n; i++) {
      ats[ords[i]] = i;
    }
    int j = 0;
    for (int i = 0; i < q; i++) {
      if (a[quer[i][0]][1] > a[quer[i][1]][1]) continue;
      for (; j < n && a[j][1] <= a[quer[i][1]][1]; j++) {
        pmx.upd(ats[j], a[j][1]);
      }
      if (quer[i][0] == quer[i][1]) {ans[quer[i][2]] = 0; continue;}
      int rb = a[quer[i][0]][1], res = 0;
      while (rb < a[quer[i][1]][0]) { // if rb < a[quer[i][1]][0], then you can't go to whatever the end is
        int k = upper_bound(s.begin(), s.end(), rb) - s.begin() - 1;
        int mx = pmx.quer(0, k);
        if (mx == rb) {
          res = -2;
          break;
        }
        rb = mx;
        res++;
      }
      res++;
      ans[quer[i][2]] = res;
    }
    for (int i = 0; i < q; i++) {
      if (ans[i] == -1) {
        cout << "impossible\n";
      } else {
        cout << ans[i] << '\n';
      }
    }
  } else {
    /*
    binary lifting probably?
    where are you going
    the thing on the right with lowest left value, hope it's this
    except you can have some annoying 2-cycles or smth
    so maybe calc dist to nearest 2cyc
    */
    vector<int> to(n, n);
    int mnl = inf, ind = n;
    for (int i = 0; i < n; i++) {
      if (mnl <= a[i][1]) {
        to[i] = ind;
      }
      if (a[i][0] < mnl) {
        a[i][0] = mnl;
        ind = i;
      }
    }
    vector<bool> cyc(n, false);
    for (int i = 1; i < n; i++) {
      if (a[i][1] == a[i - 1][1]) {
        to[i - 1] = i;
        to[i] = i - 1;
        cyc[i - 1] = cyc[i] = true;
        if (i > 2) assert(a[i - 2][1] != a[i][1]);
      }
    }
    vector<vector<int>> lift(n, vector<int>(20, 0));
    for (int i = 0; i < n; i++) {
      lift[i][0] = to[i];
    }
    for (int j = 1; j < 20; j++) {
      for (int i = 0; i < n; i++) {
        lift[i][j] = (lift[i][j - 1] == n ? n : lift[lift[i][j - 1]][j - 1]);
      }
    }
    while (q--) {
      int l, r;
      cin >> l >> r;
      l--; r--;
      l = at[l]; r = at[r];
      int ans = 0;
      for (int j = 19; j >= 0; j--) {
        if (lift[l][j] <= n && !cyc[lift[l][j]] && lift[l][j] <= r) {
          l = lift[l][j];
          ans += (1 << j);
        }
      }
      if (cyc[r]) {
        for (int i = 0; i < 2; i++) {
          l = to[l];
          ans++;
          if (l == r) break;
        }
      }
      if (l != r) {
        cout << "impossible\n";
      } else {
        cout << ans << '\n';
      }
    }
  }
}

Subtask #1
0 / 10.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 348 KB Output is correct
2 Incorrect 60 ms 15180 KB Output isn't correct
3 Halted 0 ms 0 KB -

Subtask #2
10.0 / 10.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 348 KB Output is correct
2 Correct 0 ms 348 KB Output is correct
3 Correct 8 ms 348 KB Output is correct
4 Correct 1 ms 348 KB Output is correct
5 Correct 2 ms 344 KB Output is correct
6 Correct 1 ms 348 KB Output is correct
7 Correct 1 ms 344 KB Output is correct
8 Correct 1 ms 348 KB Output is correct
9 Correct 0 ms 348 KB Output is correct

Subtask #3
0 / 15.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 348 KB Output is correct
2 Correct 0 ms 348 KB Output is correct
3 Correct 8 ms 348 KB Output is correct
4 Correct 1 ms 348 KB Output is correct
5 Correct 2 ms 344 KB Output is correct
6 Correct 1 ms 348 KB Output is correct
7 Correct 1 ms 344 KB Output is correct
8 Correct 1 ms 348 KB Output is correct
9 Correct 0 ms 348 KB Output is correct
10 Correct 1 ms 348 KB Output is correct
11 Correct 0 ms 348 KB Output is correct
12 Correct 7 ms 348 KB Output is correct
13 Correct 1 ms 348 KB Output is correct
14 Correct 2 ms 348 KB Output is correct
15 Correct 1 ms 348 KB Output is correct
16 Correct 1 ms 348 KB Output is correct
17 Correct 0 ms 348 KB Output is correct
18 Correct 0 ms 348 KB Output is correct
19 Incorrect 18 ms 2044 KB Output isn't correct
20 Halted 0 ms 0 KB -

Subtask #4
15.0 / 15.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 348 KB Output is correct
2 Correct 0 ms 348 KB Output is correct
3 Correct 8 ms 348 KB Output is correct
4 Correct 1 ms 348 KB Output is correct
5 Correct 2 ms 344 KB Output is correct
6 Correct 1 ms 348 KB Output is correct
7 Correct 1 ms 344 KB Output is correct
8 Correct 1 ms 348 KB Output is correct
9 Correct 0 ms 348 KB Output is correct
10 Correct 0 ms 348 KB Output is correct
11 Correct 0 ms 348 KB Output is correct
12 Correct 7 ms 348 KB Output is correct
13 Correct 1 ms 348 KB Output is correct
14 Correct 1 ms 348 KB Output is correct
15 Correct 1 ms 348 KB Output is correct
16 Correct 1 ms 348 KB Output is correct
17 Correct 1 ms 348 KB Output is correct
18 Correct 0 ms 348 KB Output is correct
19 Correct 1032 ms 4236 KB Output is correct
20 Correct 229 ms 3928 KB Output is correct
21 Correct 39 ms 3920 KB Output is correct
22 Correct 96 ms 3920 KB Output is correct
23 Correct 29 ms 3920 KB Output is correct
24 Correct 34 ms 3932 KB Output is correct
25 Correct 20 ms 3932 KB Output is correct

Subtask #5
0 / 20.0

# Verdict Execution time Memory Grader output
1 Incorrect 53 ms 15184 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #6
0 / 30.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 348 KB Output is correct
2 Incorrect 60 ms 15180 KB Output isn't correct
3 Halted 0 ms 0 KB -