【Cf #449 C】Willem, Chtholly and Seniorious（set维护线段）

这里介绍以个小$trick$，民间流传为$Old Driver Tree$，实质上就是$set$维护线段。

我们将所有连续一段权值相同的序列合并成一条线段，扔到$set$里去，于是$set$里的所有线段的并就是原序列，并且都不相交。

我们在操作的时候很暴力，每次把$[l, r]$的线段抠出来，暴力枚举一遍算答案。对于每一个非区间赋值的操作，最多断两条线段，新加两条线段。

实现起来很方便，思路也非常简单，但是局限性也很明显，因为其复杂度是基于随机的，并且必须存在区间赋值的操作。

但$set$维护线段的技巧还是很常见的，我用$set$和$map$都实现的一遍，发现这里用$map$相当好写。

$set$的实现：

#include <set>
#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long LL;
const int N = 100005;

int n, m, vmax, tp;
pair<LL, int> st[N];

struct Seg {
  int l, r; LL v;
  friend bool operator < (Seg a, Seg b) {
    return (a.l != b.l)? (a.l < b.l) : (a.r < b.r);
  }
};
multiset<Seg> S;
typedef multiset<Seg>::iterator Saber;

LL Pow(LL x, int b, int p, LL re = 1) {
  for (x %= p; b; b >>= 1, x = x * x % p)
    if (b & 1) re = re * x % p;
  return re;
}

namespace R {
  const int mod = 1e9 + 7;
  int seed, ret;
  int rnd() {
    ret = seed;
    seed = (seed * 7LL + 13) % mod;
    return ret;
  }
}

int main() {
  scanf("%d%d%d%d", &n, &m, &R::seed, &vmax);
  for (int i = 1; i <= n; ++i)
    S.insert((Seg){ i, i, R::rnd() % vmax + 1 });
  for (int op, l, r, x, y; m; --m) {
    op = (R::rnd() & 3) + 1;
    l = R::rnd() % n + 1;
    r = R::rnd() % n + 1;
    if (l > r) swap(l, r);
    if (op == 3) x = R::rnd() % (r - l + 1) + 1;
    else x = R::rnd() % vmax + 1;
    if (op == 4) y = R::rnd() % vmax + 1;

    Saber p = --S.lower_bound((Seg){ l + 1, 0, 0 });
    Saber q = --S.lower_bound((Seg){ r + 1, 0, 0 });
    if ((*p).l < l) S.insert((Seg){ (*p).l, l - 1, (*p).v });
    if (r < (*q).r) S.insert((Seg){ r + 1, (*q).r, (*q).v });
    if (op == 2) {
      for (Saber z = p, rin; z != q; ) {
        rin = z; ++z;
        S.erase(rin);
      }
      S.erase(q);
      S.insert((Seg){ l, r, x });
      continue;
    }
    Saber np, nq;
    if (p == q) {
      np = nq = S.insert((Seg){ l, r, (*p).v });
      S.erase(p);
    } else {
      np = S.insert((Seg){ l, (*p).r, (*p).v });
      nq = S.insert((Seg){ (*q).l, r, (*q).v });
      S.erase(p); S.erase(q);
    }
    if (op == 1) {
      Seg ins;
      for (Saber z = np, rin; z != nq; ) {
        rin = z; ++z;
        ins = *rin; ins.v += x;
        S.erase(rin); S.insert(ins);
      }
      ins = *nq; ins.v += x;
      S.erase(nq); S.insert(ins);
    }
    if (op == 3) {
      tp = 0;
      for (Saber z = np; z != nq; ++z)
        st[++tp] = make_pair((*z).v, (*z).r - (*z).l + 1);
      st[++tp] = make_pair((*nq).v, (*nq).r - (*nq).l + 1);
      sort(st + 1, st + 1 + tp);
      for (int i = 1; i <= tp; x -= st[i].second, ++i) {
        if (st[i].second >= x) {
          printf("%lld
", st[i].first);
          x = 0;
          break;
        }
      }
      if (x > 0) {
        puts("I love you, love you forever.");
        return 0;
      }
    }
    if (op == 4) {
      int ans = 0;
      for (Saber z = np; z != nq; ++z)
        ans = (ans + Pow((*z).v, x, y) * ((*z).r - (*z).l + 1)) % y;
      ans = (ans + Pow((*nq).v, x, y) * ((*nq).r - (*nq).l + 1)) % y;
      printf("%d
", ans);
    }
  }
  
  return 0;
}

$map$的实现：

#include <map>
#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long LL;
const int N = 100005;

int n, m, vmax, tp;
pair<LL, int> st[N];
map<int, LL> M;

LL Pow(LL x, int b, int p, LL re = 1) {
  for (x %= p; b; b >>= 1, x = x * x % p)
    if (b & 1) re = re * x % p;
  return re;
}

namespace R {
  const int mod = 1e9 + 7;
  int seed, ret;
  int rnd() {
    ret = seed;
    seed = (seed * 7LL + 13) % mod;
    return ret;
  }
}

int main() {
  scanf("%d%d%d%d", &n, &m, &R::seed, &vmax);
  for (int i = 1; i <= n; ++i)
    M[i] = R::rnd() % vmax + 1;
  M[n + 1] = 0;
  for (int op, l, r, x, y; m; --m) {
    op = (R::rnd() & 3) + 1;
    l = R::rnd() % n + 1;
    r = R::rnd() % n + 1;
    if (l > r) swap(l, r);
    if (op == 3) x = R::rnd() % (r - l + 1) + 1;
    else x = R::rnd() % vmax + 1;
    if (op == 4) y = R::rnd() % vmax + 1;

    auto p = --M.upper_bound(l);
    auto q = M.upper_bound(r);
    if (p->first < l) M[l] = p->second, ++p;
    if (r + 1 < q->first) --q, M[r + 1] = q->second, ++q;
    if (op == 1) {
      for (; p != q; ++p) p->second += x;
    }
    if (op == 2) {
      while (p != q) M.erase(p++);
      M[l] = x;
    }
    if (op == 3) {
      tp = 0;
      for (auto rin = p; p != q; ++p) {
        st[++tp] = { p->second, (++rin)->first - p->first };
      }
      sort(st + 1, st + 1 + tp);
      for (int i = 1; i <= tp; x -= st[i].second, ++i) {
        if (x <= st[i].second) {
          printf("%lld
", st[i].first);
          break;
        }
      }
    }
    if (op == 4) {
      int ans = 0;
      for (auto rin = p; p != q; ++p) {
        ans = (ans + Pow(p->second, x, y) * ((++rin)->first - p->first)) % y;
      }
      printf("%d
", ans);
    }
  }
  
  return 0;
}