牛客-小V和gcd树

题目传送门

sol:树剖解决，我们只维护每个节点和重儿子的边权，那么当一个节点权值改变时，也只需要修改该节点和其重儿子的边权，若该节点是其父亲节点的重儿子，则多修改一条该节点和其父亲节点的边权。查询操作时，若该节点不是父亲节点的重儿子，则暴力计算一次该节点和父亲节点的边权。

• 树链剖分

  #include <bits/stdc++.h>
  using namespace std;
  typedef long long LL;
  typedef pair<int, int> PII;
  const int MAXN = 20010;
  inline int read() {
      int n = 0, f = 1; char c = getchar();
      while (c < '0' || c > '9') {
          if (c == '-') f = -f;
          c = getchar();
      }
      while (c >= '0' && c <= '9') {
          n = 10 * n + (c ^ '0');
          c = getchar();
      }
      return f * n;
  }
  struct {
      int v, to;
  } edge[2 * MAXN];
  int head[MAXN];
  void add_edge(int u, int v) {
      static int tot = 0;
      edge[tot].v = v;
      edge[tot].to = head[u];
      head[u] = tot ++;
  }
  int fat[MAXN], son[MAXN], siz[MAXN], val[MAXN];
  void dfs1(int u, int f) {
      fat[u] = f, siz[u] = 1;
      for (int i = head[u]; i != -1; i = edge[i].to) {
          int v = edge[i].v;
          if (v == f) continue;
          dfs1(v, u);
          siz[u] += siz[v];
          if (siz[v] > siz[son[u]]) son[u] = v;
      }
  }
  int gcd[MAXN], dep[MAXN], top[MAXN], dfn[MAXN];
  void dfs2(int u, int t) {
      static int tot = 0;
      dfn[u] = ++ tot;
      dep[u] = dep[fat[u]] + 1;
      top[u] = t;
      if (t != u) gcd[dfn[u]] = __gcd(val[u], val[fat[u]]);
      if (son[u]) dfs2(son[u], t);
      for (int i = head[u]; i != -1; i = edge[i].to) {
          int v = edge[i].v;
          if (v == fat[u] || v == son[u]) continue;
          dfs2(v, v);
      }
  }
  int query(int u, int v, int k) {
      int ans = 0;
      while (top[u] != top[v]) {
          if (dep[top[u]] > dep[top[v]]) swap(u, v);
          for (int i = dfn[top[v]] + 1; i <= dfn[v]; i++)
              ans += gcd[i] <= k;
          ans += __gcd(val[top[v]], val[fat[top[v]]]) <= k;
          v = fat[top[v]];
      }
      if (dep[u] > dep[v]) swap(u, v);
      for (int i = dfn[u] + 1; i <= dfn[v]; i++)
              ans += gcd[i] <= k;
      return ans;
  }
  int main() {
      int n = read(), m = read();
      memset(head, -1, sizeof(head));
      for (int i = 1; i <= n; i++) val[i] = read();
      for (int i = 2; i <= n; i++) {
          int u = read(), v = read();
          add_edge(u, v), add_edge(v, u);
      }
      dfs1(1, 0), dfs2(1, 1);
      for (int i = 1; i <= m; i++) {
          if (read() == 1) {
              int u = read(), k = read();
              val[u] = k;
              if (top[u] != u) {
                  gcd[dfn[u]] = __gcd(val[u], val[fat[u]]);
              }
              if (son[u]) {
                  gcd[dfn[son[u]]] = __gcd(val[u], val[son[u]]);
              }
          } else {
              int u = read(), v = read(), k = read();
              printf("%d
  ", query(u, v, k));
          }
      }
      return 0;
  }

  ------------------------------------------------------------分隔线------------------------------------------------------------

  总结：$son，siz$必须在$dfs1$里求出；$dfn，top$必须在$dfs2$里求出；这次的错误是把$dfn$写到$dfs1$里面了，但是两次深搜的顺序是不一样的。第一次深搜就是按边的输入顺序进行搜索，第二次深搜在此基础上还优先搜索了重儿子。也正是因为优先搜索重儿子，才能保证重链的$dfs$序是连续的。