bzoj4551 [TJOI2016&HEOI2016]树

bzoj4551 [TJOI2016&HEOI2016]树

给定一棵有根树，请支持一下两种操作：

• 标记一个点
• 询问一个点最近一个打了标记的祖先

(n, mleq10^5)

线段树

---

首先用 (dfs) 序转换为序列问题，用线段树动态维护

打标记按照原树 (dep) 更新答案

实现时可以将点上的答案与 (tag) 合并

时间复杂度 (O(nlog n))

代码

#include <bits/stdc++.h>
using namespace std;

#define mid ((s + t) >> 1)
#define lson k << 1, s, mid
#define rson k << 1 | 1, mid + 1, t
const int maxn = 1e5 + 10;
int val[maxn << 2];
int n, m, sz[maxn], dep[maxn], tid[maxn];

vector <int> e[maxn];

int dfs(int u, int f) {
  static int now;
  tid[u] = ++now;
  int tmp = int(e[u].size());
  for (int i = 0; i < tmp; i++) {
    int v = e[u][i];
    if (v != f) {
      dep[v] = dep[u] + 1;
      sz[u] += dfs(v, u);
    }
  }
  return ++sz[u];
}

void pushtag(int k, int x) {
  if (dep[val[k]] < dep[x]) {
    val[k] = x;
  }
}

void pushdown(int k) {
  pushtag(k << 1, val[k]);
  pushtag(k << 1 | 1, val[k]);
}

void build(int k, int s, int t) {
  if (s == t) {
    val[k] = 1; return;
  }
  build(lson), build(rson);
}

void upd(int k, int s, int t, int l, int r, int x) {
  if (dep[x] < dep[val[k]]) return;
  if (l <= s && t <= r) {
    pushtag(k, x); return;
  }
  pushdown(k);
  if (l <= mid) upd(lson, l, r, x);
  if (r > mid) upd(rson, l, r, x);
}

int query(int pos) {
  int k = 1, s = 1, t = n;
  while (s < t) {
    pushdown(k);
    if (pos <= mid) {
      k <<= 1, t = mid;
    } else {
      k = k << 1 | 1, s = mid + 1;
    }
  }
  return val[k];
}

int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &u, &v);
    e[u].push_back(v), e[v].push_back(u);
  }
  dfs(1, 0), build(1, 1, n);
  char c; int u;
  while (m--) {
    scanf("%s %d", &c, &u);
    if (c == 'C') {
      upd(1, 1, n, tid[u], tid[u] + sz[u] - 1, u);
    } else {
      printf("%d
", query(tid[u]));
    }
  }
  return 0;
}