Luogu3605 [USACO17JAN]Promotion Counting晋升者计数

\(\verb|Luogu3605 [USACO17JAN]Promotion Counting晋升者计数|\)

给一棵 \(n\) 个点的树，点 \(i\) 有一个权值 \(a_i\) 。对于每个 \(i\) ，求 \(\displaystyle\sum_{j\in subtree(i)}{[a_i<a_j]}\)

\(n\leq10^5,\ fa_i<i\)

树状数组

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树上逆序对？一眼线段树合并、、、空间没毛病、、

回想求序列逆序对的过程，发现在树上做时只需减去树状数组中以往的贡献，于是便可以愉快地树状数组搞了

时间复杂度 \(O(n\log n)\) ，空间复杂度 \(O(n)\)

#include <bits/stdc++.h>
using namespace std;

#define nc getchar()
const int maxn = 1e5 + 10;
int n, a[maxn], dat[maxn], c[maxn], ans[maxn]; vector <int> e[maxn];

inline int read() {
  int x = 0; char c = nc;
  while (c < 48) c = nc;
  while (c > 47) x = x * 10 + c - 48, c = nc;
  return x;
}

void upd(int pos) {
  for (; pos <= n; pos += pos & -pos) {
    c[pos]++;
  }
}

int query(int pos) {
  int res = 0;
  for (; pos; pos &= pos - 1) {
    res += c[pos];
  }
  return res;
}

void dfs(int u) {
  int lst = query(n) - query(a[u]); upd(a[u]);
  for (int v : e[u]) dfs(v);
  ans[u] = query(n) - query(a[u]) - lst;
}

int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    dat[i] = a[i] = read();
  }
  for (int i = 2; i <= n; i++) {
    e[read()].push_back(i);
  }
  sort(dat + 1, dat + n + 1);
  for (int i = 1; i <= n; i++) {
    a[i] = lower_bound(dat + 1, dat + n + 1, a[i]) - dat;
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}