【GDKOI2014模拟】树的直径

题目大意：

题解：

#include<cstdio>
#include<algorithm>
#define fo(i, x, y) for(int i = x; i <= y; i ++)
#define fd(i, x, y) for(int i = x; i >= y; i --)
#define max(a, b) ((a) > (b) ? (a) : (b))
using namespace std;

const int N = 500000;

int n, a
, tot;
int son
[2], dep
;
int f[17]
, ans;

int lca(int x, int y) {
if(dep[x] < dep[y]) swap(x, y);
fd(i, 16, 0) if(dep[f[i][x]] >= dep[y]) x = f[i][x];
if(x == y) return x;
fd(i, 16, 0) if(f[i][x] != f[i][y]) x = f[i][x], y = f[i][y];
return f[0][x];
}

int dis(int x, int y) {
int lc = lca(x, y);
return dep[x] + dep[y] - 2 * dep[lc];
}

int main() {
scanf("%d", &n);
tot = 4;
dep[1] = 1; dep[2] = dep[3] = dep[4] = 2;
f[0][2] = f[0][3] = f[0][4] = 1;
int p = 2, q = 4;
fo(i, 1, n) {
scanf("%d", &a[i]);
son[a[i]][0] = ++ tot;
son[a[i]][1] = ++ tot;
dep[tot - 1] = dep[tot] = dep[a[i]] + 1;
f[0][tot - 1] = f[0][tot] = a[i];
fo(j, 1, 16) f[j][tot - 1] = f[j - 1][f[j - 1][tot - 1]], f[j][tot] = f[j - 1][f[j - 1][tot]];
if(dis(p, tot) < dis(q, tot)) swap(p, q);
if(dis(p, q) < dis(p, tot)) q = tot;
ans = max(ans, dis(p, q));
printf("%d\n", ans);
}
}