【BZOJ3506】【CQOI2014】排序机械臂

【题目链接】

• 点击打开链接

【双倍经验链接】

• 【BZOJ1552】【Cerc2007】robotic sort

【思路要点】

• 将权值当做下标，我们需要实现的是查询一个节点在数组中的排名，以及翻转数组的一个区间。
• 用Splay维护即可，时间复杂度$O(NLogN)$。

【代码】

#include<bits/stdc++.h>

using namespace std;
const int MAXN = 100005;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
  x = 0; int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T> void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
  write(x);
  puts("");
}
struct Splay {
  struct Node {
      int child[2];
      int father;
      int size;
      bool rev;
  } a[MAXN];
  int root, n;
  void update(int root) {
      a[root].size = 1;
      if (a[root].child[0]) a[root].size += a[a[root].child[0]].size;
      if (a[root].child[1]) a[root].size += a[a[root].child[1]].size;
  }
  void build(int &root, int l, int r, int *pos) {
      int mid = (l + r) / 2;
      root = pos[mid];
      if (l < mid) {
          build(a[root].child[0], l, mid - 1, pos);
          a[a[root].child[0]].father = root;
      }
      if (mid < r) {
          build(a[root].child[1], mid + 1, r, pos);
          a[a[root].child[1]].father = root;
      }
      update(root);
  }
  void init(int x, int *a) {
      n = x;
      a[0] = n + 1;
      a[n + 1] = n + 2;
      build(root, 0, n + 1, a);
  }
  void pushdown(int root) {
      if (a[root].rev) {
          a[root].rev = false;
          swap(a[root].child[0], a[root].child[1]);
          if (a[root].child[0]) a[a[root].child[0]].rev ^= true;
          if (a[root].child[1]) a[a[root].child[1]].rev ^= true;
      }
  }
  bool get(int x) {
      return a[a[x].father].child[1] == x;
  }
  void rotate(int x) {
      int f = a[x].father, g = a[f].father;
      pushdown(f), pushdown(x);
      bool tmp = get(x);
      a[f].child[tmp] = a[x].child[tmp ^ 1];
      if (a[f].child[tmp]) a[a[f].child[tmp]].father = f;
      a[x].child[tmp ^ 1] = f;
      a[f].father = x;
      a[x].father = g;
      if (g) a[g].child[a[g].child[1] == f] = x;
      update(f), update(x);
  }
  void splay(int x) {
      pushdown(x);
      for (int f = a[x].father; (f = a[x].father) != 0; rotate(x))
          if (a[f].father != 0) {
              if (get(f) == get(x)) rotate(f);
              else rotate(x);
          }
      root = x;
  }
  void spaly(int x) {
      pushdown(x);
      for (int f = a[x].father; (f = a[x].father) != root; rotate(x))
          if (a[f].father != root) {
              if (get(f) == get(x)) rotate(f);
              else rotate(x);
          }
  }
  int succ() {
      int now = a[root].child[1];
      pushdown(now);
      while (a[now].child[0]) {
          now = a[now].child[0];
          pushdown(now);
      }
      return now;
  }
  void work() {
      for (int i = 1; i <= n; i++) {
          splay(i);
          printf("%d", a[a[root].child[0]].size);
          if (i == n) putchar('\n');
          else putchar(' ');
          int tmp = succ();
          if (i == 1) splay(n + 1);
          else splay(i - 1);
          spaly(tmp);
          a[a[a[root].child[1]].child[0]].rev ^= true;
      }
  }
} Splay;
int n, val[MAXN], p[MAXN];
bool cmp(int x, int y) {
  if (val[x] == val[y]) return x < y;
  else return val[x] < val[y];
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++)
      read(val[i]), p[i] = i;
  sort(p + 1, p + n + 1, cmp);
  for (int i = 1; i <= n; i++)
      val[p[i]] = i;
  Splay.init(n, val);
  Splay.work();
  return 0;
}