给一棵边带权树，问两点之间的距离小于等于K的点对有多少个。
树上的点 分治，目前还不懂。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(x,v) memset(x,v,sizeof(x))
#define go(i,a,b) for (int i = a; i <= b; i++)
#define og(i,a,b) for (int i = a; i >= b; i--)
using namespace std;
typedef long long LL;
const double EPS = 1e-10;
const int INF = 0x3f3f3f3f;
const int N = 1e4+10;
struct node{
int val,v;
};
vector<node>g[N];
int n,size,s[N],f[N],root,d[N],K,ans;
vector<int>dep;
bool vis[N];
void getroot(int now, int fa){
s[now] = 1; f[now] = 0; int u;
for (int i = 0; i < g[now].size(); i++)
if ((u = g[now][i].v) != fa && !vis[u]){
getroot(u,now);
s[now] += s[u];
f[now] = max(f[now],s[u]);
}
f[now] = max(f[now],size - s[now]);
if (f[now] < f[root]) root = now;
}
void getdep(int now, int fa){
int u; dep.push_back(d[now]);
s[now] = 1;
for (int i = 0; i < g[now].size(); i++)
if ((u = g[now][i].v) != fa && !vis[u]){
d[u] = d[now] + g[now][i].val;
getdep(u,now);
s[now] += s[u];
}
}
int calc(int now, int op){
dep.clear(); d[now] = op;
getdep(now,0);
sort(dep.begin(),dep.end());
int ret = 0;
for (int l = 0, r = dep.size() - 1; l < r;)
if (dep[l] + dep[r] <= K) ret += r-l++; else
r--;
return ret;
}
void work(int now){
int u;
ans += calc(now,0); vis[now] = 1;
for(int i = 0; i < g[now].size(); i++)
if (!vis[u = g[now][i].v]){
ans -= calc(u,g[now][i].val);
f[0] = size = s[u];
getroot(u,root = 0);
work(root);
}
}
int main(){
while(scanf("%d%d",&n,&K) == 2 && n){
go(i,0,n) g[i].clear();
mem(vis,0);
int x,y,z;
go(i,1,n-1){
scanf("%d%d%d",&x,&y,&z);
node e; e.v = y, e.val = z;
g[x].push_back(e); e.v = x;
g[y].push_back(e);
}
f[0] = size = n;
getroot(1,root = 0);
ans = 0;
work(root);
printf("%d\n",ans);
}
return 0;
}