题意
给定一棵树，求每一棵子树内距离跟最小的节点数最多的深度。
\(n \le 1e6\)
Sol
dsu 板子。
我们先考虑那个 \(n ^ 2\) 的 dp。
对于每一个节点 \(x\)，用 \(f_i\) 表示当前在 \(x\) 子树内深度为 \(i\) 的节点有多少个。
求最大值用一个变量 \(tot\) 表示即可。
考虑直接上启发式合并即可。
Code
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <queue>
using namespace std;
#ifdef ONLINE_JUDGE
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;
#endif
int read() {
int p = 0, flg = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') flg = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
p = p * 10 + c - '0';
c = getchar();
}
return p * flg;
}
void write(int x) {
if (x < 0) {
x = -x;
putchar('-');
}
if (x > 9) {
write(x / 10);
}
putchar(x % 10 + '0');
}
const int N = 1e6 + 5, M = 2e6 + 5;
namespace G {
array <int, N> fir;
array <int, M> nex, to;
int cnt;
void add(int x, int y) {
cnt++;
nex[cnt] = fir[x];
to[cnt] = y;
fir[x] = cnt;
}
}
array <int, N> ans;
namespace Hpt {
using G::fir; using G::nex; using G::to;
array <int, N> son, fa, dep, siz;
array <int, N> dfn, idx;
int cnt;
void dfs1(int x) {
cnt++;
dfn[x] = cnt;
idx[cnt] = x;
siz[x] = 1;
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == fa[x]) continue;
fa[to[i]] = x;
dep[to[i]] = dep[x] + 1;
dfs1(to[i]);
siz[x] += siz[to[i]];
if (siz[to[i]] > siz[son[x]]) son[x] = to[i];
}
}
queue <int> q;
array <int, N> cur;
int tot;
void add(int x) {
q.push(x);
cur[x]++;
if (cur[x] > cur[tot]) tot = x;
else if (cur[x] == cur[tot]) tot = min(tot, x);
}
void clear() {
while (!q.empty())
cur[q.front()] = 0, q.pop();
/* tot = 1e6 + 1; */
}
void dfs2(int x) {
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == fa[x] || to[i] == son[x]) continue;
dfs2(to[i]); clear();
}
tot = dep[x];
if (son[x]) dfs2(son[x]);
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == fa[x] || to[i] == son[x]) continue;
for (int j = dfn[to[i]]; j <= dfn[to[i]] + siz[to[i]] - 1; j++)
add(dep[idx[j]]);
}
add(dep[x]);
ans[x] = tot - dep[x];
}
}
int main() {
int n = read();
for (int i = 2; i <= n; i++) {
int x = read(), y = read();
G::add(x, y), G::add(y, x);
}
/* Hpt::tot = 1e6 + 1; */
Hpt::dfs1(1), Hpt::dfs2(1);
for (int i = 1; i <= n; i++)
write(ans[i]), puts("");
return 0;
}