Submission #980077

#	Time	Username	Problem	Language	Result	Execution time	Memory
980077		asdfgrace	Event Hopping (BOI22_events)	C++17	25 / 100	1580 ms	8656 KiB

events

#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'));

/*
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];
  });
  vector<int> at(n);
  for (int i = 0; i < n; i++) {
    at[a[i][2]] = i;
  }
  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';
    }
  }
}

• Subtask 1: 0 / 10
• Subtask 2: 10 / 10
• Subtask 3: 0 / 15
• Subtask 4: 15 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 30