Educational Codeforces Round 47 F. Dominant Indices 线段树合并/长链剖分

F. Dominant Indices

题意：给一颗以1为根的树，设$d[u][x]$为$u$子树中距离$u$为$x$的点的数量，对于每个点$u$，求最小的$x$使得$d[u][x]$最大

解法：

线段树合并：对于每个子树，用一颗线段树维护其$d[u][x]$的最大值，然后向上走不停的线段树合并即可。

长链剖分：对于每个子树，用一个动态数组维护其$d[u][x]$所有情况，每次向上合并的时候，不停的把轻儿子的链合并到重儿子的链，每个点代表的深度的信息只会被合并一次，复杂度极其优秀

线段树合并

#include 
using namespace std;
const int maxn = 1e6 + 6;
#define mid (l + r) / 2
int rt[maxn], ls[maxn * 21], rs[maxn * 21], dep[maxn];
int mx[maxn * 21], pos[maxn * 21], ans[maxn], cnt, n;
vector<int> G[maxn];
void pushup(int o)
{
    mx[o] = max(mx[ls[o]], mx[rs[o]]);
    if (mx[ls[o]] >= mx[rs[o]])
        pos[o] = pos[ls[o]];
    else
        pos[o] = pos[rs[o]];
}
void up(int &o, int l, int r, int k, int v)
{
    if (!o)
        o = ++cnt;
    mx[o] += v;
    if (l == r)
    {
        pos[o] = l;
        return;
    }
    if (k <= mid)
        up(ls[o], l, mid, k, v);
    else
        up(rs[o], mid + 1, r, k, v);
    pushup(o);
}
int merge(int &o, int pre, int l, int r)
{
    if (!o || !pre)
        return o | pre;
    if (l == r)
    {
        mx[o] += mx[pre];
        return o;
    }
    ls[o] = merge(ls[o], ls[pre], l, mid);
    rs[o] = merge(rs[o], rs[pre], mid + 1, r);
    pushup(o);
    return o;
}
void dfs(int u, int fa)
{
    dep[u] = dep[fa] + 1;
    up(rt[u], 1, n, dep[u], 1);
    for (auto v : G[u])
        if (v != fa)
        {
            dfs(v, u);
            rt[u] = merge(rt[u], rt[v], 1, n);
        }
    ans[u] = pos[rt[u]] - dep[u];
}
int main()
{
    int u, v;
    scanf("%d", &n);
    for (int i = 1; i < n; i++)
    {
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++)
        printf("%d\n", ans[i]);
}

长链剖分

#include 
using namespace std;
const int maxn = 1e6 + 6;
vector<int> G[maxn], g[maxn];
int h[maxn], son[maxn], ans[maxn];
void dfs(int u, int fa)
{
    for (auto v : G[u])
        if (v != fa)
        {
            dfs(v, u);
            if (h[v] + 1 > h[u])
            {
                h[u] = h[v] + 1;
                son[u] = v;
            }
        }
    if (!son[u])
    {
        g[u].push_back(1);
        return;
    }
    ans[u] = ans[son[u]];
    swap(g[u], g[son[u]]);
    for (auto v : G[u])
        if (v != fa && v != son[u])
        {
            int n = g[u].size();
            int m = g[v].size();
            for (int i = n - m; i < n; i++)
            {
                g[u][i] += g[v][i - (n - m)];
                if (g[u][i] >= g[u][ans[u]])
                    ans[u] = i;
            }
        }
    g[u].push_back(1);
    if (g[u][ans[u]] == 1)
        ans[u] = g[u].size() - 1;
}
int main()
{
    int n, u, v;
    scanf("%d", &n);
    for (int i = 1; i < n; i++)
    {
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++)
        printf("%d\n", h[i] - ans[i]);
}