题解：lca预处理，树上差分套线段树合并复杂度O（nlogn）但是跑起来比树链剖分（nlognlogn）的还慢...........
#include"bits/stdc++.h"
#define lson l,mid,ls[rt]
#define rson mid+1,r,rs[rt]
using namespace std;
typedef long long LL;
const int MX = 1e6+7;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
int mm,n,m,K;
vector<int> add[MX],sub[MX];
struct node{
int u,v,c;
}p[MX];
vector<int> G[MX];
int f[MX][22], ans[MX],deep[MX];
void dfs(int u,int fa)
{
f[u][0] = fa;
deep[u] = deep[f[u][0]] + 1;
for(int i = 0; f[u][i]; i++){
f[u][i+1] = f[f[u][i]][i];
}
for(auto v : G[u]){
if(v != fa) dfs(v,u);
}
}
int lca(int x, int y)
{
if(deep[x] < deep[y]) swap(x, y);
for(int i = 20; i >= 0; i--)
if(deep[f[x][i]] >= deep[y])
x = f[x][i];
if(x == y) return x;
for(int i = 20; i >= 0; i--)
if(f[x][i] != f[y][i])
x = f[x][i], y = f[y][i];
return f[x][0];
}
int sum[MX*20],ls[MX*20],rs[MX*20],now,rot[MX];
void push_up(int rt)
{
sum[rt] = sum[ls[rt]] + sum[rs[rt]];
}
void update(int x, int c, int l, int r, int &rt)
{
if(!rt) rt = ++now;
assert(now < MX*20 && rt > 0);
if(l == r){
sum[rt] = c;
return;
}
int mid = (l+r)>>1;
if(x <= mid) update(x,c,lson);
else update(x,c,rson);
push_up(rt);
}
int query(int k, int l, int r, int rt)
{
if(!rt) return mm;
if(l == r) return l;
int mid = (l+r)>>1;
int ret = mm;
if(sum[ls[rt]] >= k) ret = query(k,lson);
else ret = query(k-sum[ls[rt]],rson);
return ret;
}
int merg(int rt1, int rt2, int l, int r){
if(!rt2) return rt1;
if(!rt1) return rt2;
if(l==r) return rt1;
int mid = (l+r)>>1;
ls[rt1] = merg(ls[rt1],ls[rt2],l,mid);
rs[rt1] = merg(rs[rt1],rs[rt2],mid+1,r);
push_up(rt1);
return rt1;
}
void work(int u, int fa)
{
for(auto v : G[u]){
if(v == fa) continue;
work(v,u);
merg(rot[u],rot[v],1,mm);
}
for(auto id : add[u])
update(id,1,1,mm,rot[u]);
int id = m+1-query(K,1,mm,rot[u]);
ans[u] = p[id].c;
for(auto id : sub[u])
update(id,0,1,mm,rot[u]);
}
int main()
{
scanf("%d%d%d",&n,&m,&K);
assert(K >= 1);
for(int i = 1,u,v; 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 <= m; i++) {
scanf("%d%d%d",&p[i].u,&p[i].v,&p[i].c);
}
for(int i = m; i >= 1; i--) {
int u = p[i].u, v = p[i].v, c = p[i].c;
int rt = lca(u,v);
sub[rt].push_back(m-i+1);
if(rt != u) add[u].push_back(m-i+1);
if(rt != v) add[v].push_back(m-i+1);
if(u == v) add[u].push_back(m-i+1);
}
for(int i = 1; i <= n; i++) rot[i] = i;
now = n;
mm = m+1;
work(1,0);
for(int i = 1; i <= n; i++)
printf("%d%c",ans[i]," \n"[i==n]);
return 0;
}