题面

思路

本蒟蒻第一次学点分治，正遇模板题，留个模板代码

\(Code\)

#include<cstdio>
#include<algorithm>
using namespace std;

const int N = 1e4 + 5;
int len , d[N] , cnt , n , use[N] , h[N] , tot , size , son[N] , siz[N] , rt , ans;

struct edge{
	int to , nxt , w;
}e[2 * N];

inline void add(int x , int y , int z)
{
	e[++tot].to = y;
	e[tot].w = z;
	e[tot].nxt = h[x];
	h[x] = tot;
}

inline void getrt(int x , int fa)
{
	son[x] = 0 , siz[x] = 1;
	for(register int i = h[x]; i; i = e[i].nxt)
	{
		int v = e[i].to;
		if (use[v] || v == fa) continue;
		getrt(v , x);
		siz[x] += siz[v];
		son[x] = max(son[x] , siz[v]);
	}
	son[x] = max(son[x] , size - siz[x]);
	rt = son[x] < son[rt] ? x : rt;
}

inline void getdis(int x , int fa , int dis)
{
	for(register int i = h[x]; i; i = e[i].nxt)
	{
		int v = e[i].to;
		if (use[v] || v == fa) continue;
		d[++cnt] = dis + e[i].w;
		getdis(v , x , d[cnt]);
	}
}

inline int binary(int l , int r , int x)
{
	int mid , res = 0;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (d[mid] <= x) res = mid , l = mid + 1;
		else r = mid - 1;
	}
	return res;
}

inline int getans(int x , int dis)
{
	d[cnt = 1] = dis;
	getdis(x , 0 , dis);
	sort(d + 1 , d + cnt + 1);
	int l = 1 , r , res = 0;
	while (len - d[l] >= d[l] && l < cnt)
	{
		r = binary(l + 1 , cnt , len - d[l]);
		if (r > l) res += r - l;
		l++;
	}
	return res;
}

inline void divide(int x)
{
	use[x] = 1 , ans += getans(x , 0);
	for(register int i = h[x]; i; i = e[i].nxt)
	{
		int v = e[i].to;
		if (use[v]) continue;
		ans -= getans(v , e[i].w);
		size = siz[v] , rt = 0;
		getrt(v , x) , divide(rt);
	}
}

int main()
{
	scanf("%d%d" , &n , &len);
	int u , v , w;
	for(register int i = 1; i < n; i++)
	{
		scanf("%d%d%d" , &u , &v , &w);
		add(u , v , w) , add(v , u , w);
	}
	size = n;
	son[0] = 2e9;
	getrt(1 , 0) , divide(rt);
	printf("%d" , ans);
}

JZOJ 1166. 树中点对距离

原文：https://www.cnblogs.com/leiyuanze/p/13398839.html