[APIO2018] New Home new home [tree line, multiset]

Each point represents the precursor of the current segment tree point, that the first occurrence of color, just write about this stuff multiset would be finished.
It is important how the query answer, no solution is clear that first sentence out.
Binary tree line on it

#include <bits/stdc++.h>

using namespace std;
int read() {
  int x = 0;
  char c = getchar();
  while (c < 48) c = getchar();
  while (c > 47) x = x * 10 + (c - 48), c = getchar();
  return x;
}

int min(int x, int y) { return x < y ? x : y; }
int max(int x, int y) { return x > y ? x : y; }

int n, k, q;
const int maxn = 3e5 + 53;
const int maxm = 1e7;

multiset<int> st[maxn];
int ans[maxn];

struct node {
  int x, t, id, type;
  bool operator<(const node& other) const {
    if (t != other.t) return t < other.t;
    return type < other.type;
  }
} t[maxn << 2];

int R;
int rt, cnt = 0;
int ls[maxm], rs[maxm], mn[maxm];
multiset<int> ms[maxm];
void build(int& p, int l, int r) {
  p = ++cnt;
  if (l == r) {
    for (int i = 1; i <= k; i++) ms[p].insert(0);
    return;
  }
  int mid = l + r >> 1;
  build(rs[p], mid + 1, r);
}

void modify(int& p, int l, int r, const int& x, const int& inc, const int& del) {
  if (!p) p = ++cnt;
  if (l == r) {
    if (inc >= 0) ms[p].insert(inc);
    if (del >= 0) ms[p].erase(ms[p].find(del));
    mn[p] = ((ms[p].size()) ? *ms[p].begin() : R);
    return;
  }
  int mid = l + r >> 1;
  if (x <= mid)
    modify(ls[p], l, mid, x, inc, del);
  else
    modify(rs[p], mid + 1, r, x, inc, del);
  mn[p] = min(mn[ls[p]], mn[rs[p]]);
}

int qry(int x) {
  int l = 1, r = R;
  int p = rt, chk = x << 1, tmn = R;
  while (l < r) {
    int mid = l + r >> 1, d = min(tmn, mn[rs[p]]);
    if (mid < x || d + mid < chk || d < 1)
      l = mid + 1, p = rs[p];
    else
      tmn = d, r = mid, p = ls[p];
  }
  return l - x;
}

int main() {
  // freopen("testdata.in", "r", stdin);
  n = read(), k = read(), q = read();
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    int x = read(), id = read(), a = read(), b = read();
    R = max(R, x);
    t[++cnt] = { x, a, id, 0 }, t[++cnt] = { x, b + 1, id, 1 };
  }
  for (int i = 1; i <= q; i++) {
    int x = read(), time = read();
    t[++cnt] = { x, time, i, 2 };
  }

  R = R << 1 | 1, build(rt, 1, R), mn[0] = R;
  for (int i = 1; i <= k; i++) st[i].insert(0), st[i].insert(R);

  sort(t + 1, t + cnt + 1);
  int tot = 0;
  for (int i = 1; i <= cnt; i++) {
    int x = t[i].x, id = t[i].id;
    if (t[i].type == 0) {
      auto it = st[id].lower_bound(x);
      auto it2 = it;
      it2--;
      modify(rt, 1, R, x, *it2, -1);
      modify(rt, 1, R, *it, x, *it2);
      if (st[id].size() == 2) ++tot;
      st[id].insert(x);
    }
    if (t[i].type == 1) {
      st[id].erase(st[id].find(x));
      auto it = st[id].lower_bound(x);
      auto it2 = it;
      it2--;
      modify(rt, 1, R, x, -1, *it2);
      modify(rt, 1, R, *it, *it2, x);
      if (st[id].size() == 2) --tot;
    }
    if (t[i].type == 2) {
      if (tot == k)
        ans[id] = qry(x);
      else
        ans[id] = -1;
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}