P3605 Promotion Counting P

考点

  • 离散化
  • 线段树

题解

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#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 50;
// num:离散后的排名
int n, num, rt[maxn], dis[maxn], a[maxn], ans[maxn];
int tot, head[maxn], ver[maxn << 1], nxt[maxn << 1];
// 权值线段树
int cnt, lc[maxn * 18], rc[maxn * 18], s[maxn * 18];

void add(int x, int y) { ver[++tot] = y, nxt[tot] = head[x], head[x] = tot; }

void update(int &p, int l, int r, int x) {
if (!p) p = ++cnt;
if (l == r) {
s[p] += 1;
return;
}
int mid = (l + r) >> 1;
if (x <= mid) update(lc[p], l, mid, x);
if (x > mid) update(rc[p], mid + 1, r, x);
s[p] = s[lc[p]] + s[rc[p]];
}

int ask(int p, int l, int r, int L, int R) {
if (l >= L && r <= R) return s[p];
int mid = (l + r) >> 1, ans = 0;
if (L <= mid && lc[p]) ans += ask(lc[p], l, mid, L, R);
if (R > mid && rc[p]) ans += ask(rc[p], mid + 1, r, L, R);
return ans;
}

int merge(int p, int q, int l, int r) {
if (!p || !q) return p + q;
if (l == r) {
s[p] += s[q];
return p;
}
int mid = (l + r) >> 1;
lc[p] = merge(lc[p], lc[q], l, mid);
rc[p] = merge(rc[p], rc[q], mid + 1, r);
s[p] = s[lc[p]] + s[rc[p]];
return p;
}

void dfs(int x, int f) {
for (int i = head[x]; i; i = nxt[i]) {
int y = ver[i];
if (y == f) continue;
dfs(y, x);
rt[x] = merge(rt[x], rt[y], 1, num);
}
if (a[x] < num) ans[x] = ask(rt[x], 1, num, a[x] + 1, num);
}

int main() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
dis[i] = a[i];
}
sort(dis + 1, dis + 1 + n);
num = unique(dis + 1, dis + 1 + n) - 1 - dis;
for (int i = 1; i <= n; ++i) {
a[i] = lower_bound(dis + 1, dis + 1 + num, a[i]) - dis;
}
update(rt[1], 1, num, a[1]);
for (int x, i = 2; i <= n; ++i) {
scanf("%d", &x);
add(x, i), add(i, x);
update(rt[i], 1, num, a[i]);
}
dfs(1, 0);
for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
return 0;
}

思路

求每个点,其子树下有多少人的值是大于它的;直接每个节点开权值线段树,然后向上合并即可。

由于合并时会对已经合并后的子树造成影响,所以每个节点合并完成时需要即刻记录答案,而不能延后!

题目说了每个人的值都不一样,尽管值域高达1e9,但是可以通过离散化压到1e5