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 (displaystylesum_{jin subtree(i)}{[a_i<a_j]})
(nleq10^5, fa_i<i)
树状数组
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(nlog 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
", ans[i]);
}
return 0;
}