虽然是裸的换根dp，但是为了在联赛前锻炼码力，强行上了点分树+线段树.
写完+调完总共花了不到 $50$ 分钟，感觉还行.
code:
#include <bits/stdc++.h>
#define N 420004
#define LL long long
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
namespace IO
{
char *p1,*p2,buf[100000];
#define nc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
int rd() {int x=0; char c=nc(); while(c<48) c=nc(); while(c>47) x=(((x<<2)+x)<<1)+(c^48),c=nc(); return x;}
};
int n,K,edges;
int hd[N],nex[N<<1],val[N<<1],to[N<<1],F[N],G[N];
inline void add(int u,int v,int c)
{
nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;
}
struct tree
{
LL dis[N];
int size[N],son[N],top[N],fa[N],dep[N];
void dfs1(int u,int ff)
{
fa[u]=ff,dep[u]=dep[ff]+1,size[u]=1;
for(int i=hd[u];i;i=nex[i])
{
int v=to[i];
if(v==ff) continue;
dis[v]=dis[u]+val[i];
dfs1(v,u);
size[u]+=size[v];
if(size[v]>size[son[u]]) son[u]=v;
}
}
void dfs2(int u,int tp)
{
top[u]=tp;
if(son[u]) dfs2(son[u],tp);
for(int i=hd[u];i;i=nex[i])
if(to[i]!=fa[u]&&to[i]!=son[u]) dfs2(to[i],to[i]);
}
inline int LCA(int x,int y)
{
while(top[x]!=top[y])
{
dep[top[x]]>dep[top[y]]?x=fa[top[x]]:y=fa[top[y]];
}
return dep[x]<dep[y]?x:y;
}
inline LL Dis1(int x,int y)
{
return dis[x]+dis[y]-(dis[LCA(x,y)]<<1);
}
inline LL Dis2(int x,int y)
{
return dep[x]+dep[y]-(dep[LCA(x,y)]<<1);
}
}tree;
struct seg
{
int tot;
inline void init() { tot=0;}
inline int newnode() { return ++tot; }
struct node
{
int ls,rs,s2;
LL sum;
}t[N*20];
void update(int &x,int l,int r,int p,LL v)
{
if(!x) x=newnode();
t[x].sum+=v;
t[x].s2++;
if(l==r) return;
int mid=(l+r)>>1;
if(p<=mid) update(t[x].ls,l,mid,p,v);
else update(t[x].rs,mid+1,r,p,v);
}
LL query(int x,int l,int r,int L,int R)
{
if(!x) return 0;
if(l>=L&&r<=R) return t[x].sum;
int mid=(l+r)>>1;
LL re=0ll;
if(L<=mid) re+=query(t[x].ls,l,mid,L,R);
if(R>mid) re+=query(t[x].rs,mid+1,r,L,R);
return re;
}
int q2(int x,int l,int r,int L,int R)
{
if(!x) return 0;
if(l>=L&&r<=R) return t[x].s2;
int mid=(l+r)>>1,re=0;
if(L<=mid) re+=q2(t[x].ls,l,mid,L,R);
if(R>mid) re+=q2(t[x].rs,mid+1,r,L,R);
return re;
}
}seg;
struct opt
{
int root,sn;
int Fa[N],size[N],vis[N],mx[N];
void getroot(int u,int ff)
{
size[u]=1,mx[u]=0;
for(int i=hd[u];i;i=nex[i])
{
int v=to[i];
if(v==ff||vis[v]) continue;
getroot(v,u);
size[u]+=size[v];
mx[u]=max(mx[u], size[v]);
}
mx[u]=max(mx[u], sn-size[u]);
if(mx[u]<mx[root]) root=u;
}
void calc(int pp,int u,int lst,int ff,int dep,LL w)
{
if(dep<=K) seg.update(F[pp],0,K,dep,w);
if(ff)
{
LL tmp2=tree.Dis1(u,ff);
int tmp1=tree.Dis2(u,ff);
if(tmp1<=K) seg.update(G[pp],0,K,tmp1,tmp2);
}
for(int i=hd[u];i;i=nex[i])
{
int v=to[i];
if(vis[v]||v==lst) continue;
calc(pp,v,u,ff,dep+1,w+1ll*val[i]);
}
}
void dfs(int u)
{
vis[u]=1;
calc(u,u,0,Fa[u],0,0ll);
for(int i=hd[u];i;i=nex[i])
{
int v=to[i];
if(vis[v]) continue;
root=0,sn=size[v],getroot(v,u),Fa[root]=u,dfs(root);
}
}
LL getdis(int u)
{
LL tmp=seg.query(F[u],0,K,0,K);
int U=u;
for(;Fa[u];u=Fa[u])
{
int tt=tree.Dis2(U,Fa[u]);
if(tt<=K)
{
LL a=seg.query(F[Fa[u]],0,K,0,K-tt)-seg.query(G[u],0,K,0,K-tt);
LL b=seg.q2(F[Fa[u]],0,K,0,K-tt)-seg.q2(G[u],0,K,0,K-tt);
tmp+=a+b*tree.Dis1(U,Fa[u]);
}
}
return tmp;
}
void solve()
{
root=0;
mx[0]=sn=n;
getroot(1,0);
dfs(root);
int i,j;
for(i=1;i<=n;++i)
{
// getdis(i);
printf("%lld ",getdis(i));
}
}
}opt;
int main()
{
// setIO("input");
using namespace IO;
int i,j;
n=rd(),K=rd();
// scanf("%d%d",&n,&K);
for(i=1;i<n;++i)
{
int x,y,c;
x=rd(),y=rd(),c=rd();
add(x,y,c),add(y,x,c);
}
tree.dfs1(1,0);
tree.dfs2(1,1);
seg.init();
opt.solve();
return 0;
}