[Vani有约会]雨天的尾巴

嘟嘟嘟




看到链上操作，自然想到树剖。

先考虑序列上的问题：那么区间修改可以用差分。所以我们把操作拆成\(L\)和\(R + 1\)两个点，然后离线。排序后扫一遍，用线段树维护数量最多的颜色是哪一个。




现在改成了树上，那么就用树剖把链变成一段段区间，然后做法和上面就一样了。

复杂度\(O(n ^ 2logn)\)。




（据说这题还有线段树合并的做法，但是我没想出来）

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
const int maxq = 4e6 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m;
struct Edge
{
  int nxt, to;
}e[maxn << 1];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y)
{
  e[++ecnt] = (Edge){head[x], y};
  head[x] = ecnt;
}

int dep[maxn], siz[maxn], son[maxn], fa[maxn];
In void dfs1(int now, int _f)
{
  siz[now] = 1;
  for(int i = head[now], v; ~i; i = e[i].nxt)
    {
      if((v = e[i].to) == _f) continue;
      dep[v] = dep[now] + 1;
      fa[v] = now;
      dfs1(v, now);
      siz[now] += siz[v];
      if(!son[now] || siz[son[now]] < siz[v]) son[now] = v;
    }
}
int dfsx[maxn], pos[maxn], top[maxn], cnt = 0;
In void dfs2(int now, int _f)
{
  dfsx[now] = ++cnt; pos[cnt] = now;
  if(son[now]) top[son[now]] = top[now], dfs2(son[now], now);
  for(int i = head[now], v; ~i; i = e[i].nxt)
    {
      if((v = e[i].to) == _f || v == son[now]) continue;
      top[v] = v, dfs2(v, now);
    }
}
struct Node
{
  int id, flg, col;
  In bool operator < (const Node& oth)const
  {
    return id < oth.id;
  }
}q[maxq];
int qcnt = 0;
In void solve(int x, int y, int col)
{
  while(top[x] ^ top[y])
    {
      if(dep[top[x]] < dep[top[y]]) swap(x, y);
      q[++qcnt] = (Node){dfsx[top[x]], 1, col};
      q[++qcnt] = (Node){dfsx[x] + 1, -1, col};
      x = fa[top[x]];
    }
  if(dfsx[x] < dfsx[y]) swap(x, y);
  q[++qcnt] = (Node){dfsx[y], 1, col};
  q[++qcnt] = (Node){dfsx[x] + 1, -1, col};
}

int Max_col;
int l[maxn << 2], r[maxn << 2], Max[maxn << 2], Mpos[maxn << 2];
In void build(int L, int R, int now)
{
  l[now] = L; r[now] = R;
  if(L == R) {Mpos[now] = L; return;}
  int mid = (L + R) >> 1;
  build(L, mid, now << 1);
  build(mid + 1, R, now << 1 | 1);
  Mpos[now] = Mpos[now << 1];
}
In void update(int id, int now, int d)
{
  if(l[now] == r[now]) {Max[now] += d; return;}
  int mid = (l[now] + r[now]) >> 1;
  if(id <= mid) update(id, now << 1, d);
  else update(id, now << 1 | 1, d);
  if(Max[now << 1] >= Max[now << 1 | 1]) Max[now] = Max[now << 1], Mpos[now] = Mpos[now << 1];
  else Max[now] = Max[now << 1 | 1], Mpos[now] = Mpos[now << 1 | 1];
}

int ans[maxn];

int main()
{
  Mem(head, -1);
  n = read(), m = read();
  for(int i = 1; i < n; ++i)
    {
      int x = read(), y = read();
      addEdge(x, y), addEdge(y, x);
    }
  dfs1(1, 0); top[1] = 1, dfs2(1, 0);
  for(int i = 1; i <= m; ++i)
    {
      int x = read(), y = read(), col = read();
      Max_col = max(Max_col, col);
      solve(x, y, col);
    }
  sort(q + 1, q + qcnt + 1);
  build(0, Max_col, 1);
  for(int i = 1, j = 1; i <= cnt; ++i)
    {
      while(q[j].id == i) update(q[j].col, 1, q[j].flg), ++j;
      ans[pos[i]] = Mpos[1];
    }
  for(int i = 1; i <= n; ++i) write(ans[i]), enter;
  return 0;
}