感觉现在写点分治可快了~
二分答案，就可以将求第 $k$ 大转换成一个判断问题，直接拿点分树判断一下就行了.
#include <cstdio>
#include <vector>
#include <algorithm>
#define N 100004
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
int n,edges,K;
int hd[N],to[N<<1],nex[N<<1],val[N<<1];
void add(int u,int v,int c)
{
nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;
}
namespace tree
{
int size[N],son[N],fa[N],top[N],dep[N],dis[N];
void dfs1(int u,int ff)
{
fa[u]=ff,size[u]=1;
for(int i=hd[u];i;i=nex[i])
if(to[i]!=ff)
{
dep[to[i]]=dep[u]+1,dis[to[i]]=dis[u]+val[i];
dfs1(to[i],u);
size[u]+=size[to[i]];
if(size[to[i]]>size[son[u]]) son[u]=to[i];
}
}
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]);
}
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;
}
int Dis(int x,int y)
{
return dis[x]+dis[y]-(dis[LCA(x,y)]<<1);
}
};
vector<int>F[N],G[N];
int root,sn;
int mx[N],size[N],vis[N],Fa[N];
void dfs(int u,int ff)
{
size[u]=1;
for(int i=hd[u];i;i=nex[i])
if(!vis[to[i]]&&to[i]!=ff)
dfs(to[i],u),size[u]+=size[to[i]];
}
void getroot(int u,int ff)
{
size[u]=1,mx[u]=0;
for(int i=hd[u];i;i=nex[i])
if(to[i]!=ff&&!vis[to[i]])
getroot(to[i],u),size[u]+=size[to[i]],mx[u]=max(mx[u],size[to[i]]);
mx[u]=max(mx[u],sn-size[u]);
if(mx[u]<mx[root]) root=u;
}
void calc(int u,int ff,int dep,int rt)
{
F[rt].push_back(dep);
if(Fa[rt]) G[rt].push_back(tree::Dis(Fa[rt], u));
for(int i=hd[u];i;i=nex[i])
if(to[i]!=ff&&!vis[to[i]])
calc(to[i],u,dep+val[i],rt);
}
void prepare(int u)
{
vis[u]=1;
calc(u,0,0,u);
F[u].push_back(1000000004);
sort(F[u].begin(),F[u].end());
if(Fa[u])
{
G[u].push_back(1000000004);
sort(G[u].begin(),G[u].end());
}
for(int i=hd[u];i;i=nex[i])
if(!vis[to[i]])
dfs(to[i],u),sn=size[to[i]],root=0,getroot(to[i],u),Fa[root]=u,prepare(root);
}
int query(int u,int k)
{
int U=u, re=upper_bound(F[u].begin(),F[u].end(),k)-F[u].begin()-1;
for(;Fa[u];u=Fa[u])
{
int dis=tree::Dis(Fa[u],U);
if(dis<=k)
{
re+=(upper_bound(F[Fa[u]].begin(),F[Fa[u]].end(),k-dis)-F[Fa[u]].begin());
re-=(upper_bound(G[u].begin(),G[u].end(),k-dis)-G[u].begin());
}
}
return re;
}
int main()
{
int i,j,tot=0;
char ss[2];
// setIO("input");
scanf("%s%d%d",ss,&n,&K);
for(i=1;i<n;++i)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c),add(a,b,c),add(b,a,c),tot+=c;
}
tree::dfs1(1,0);
tree::dfs2(1,1);
mx[0]=sn=n,root=0,getroot(1,0),prepare(root);
for(i=1;i<=n;++i)
{
int l=1,r=tot,mid,ans=0;
while(l<=r)
{
mid=(l+r)>>1;
if(query(i,mid)>=K)
{
ans=mid,r=mid-1;
}
else l=mid+1;
}
printf("%d
",ans);
}
return 0;
}