[APIO2018] New Home 新家 [线段树，multiset]

线段树的每个点表示当前点的前驱，即这个颜色上一次出现的位置，这个玩意multiset随便写写就完了。
重要的是怎么查询答案，无解显然先判掉。
线段树上二分就可以了

#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;
}