AcWing 252 树 (点分治)

题目链接：https://www.acwing.com/problem/content/description/254/

每次找到树的重心，分治下去统计答案（经过当前根节点的路径）即可

统计答案使用了指针扫描数组的方法，要注意去掉同一子树内路径的答案
还可以直接在树上统计子树答案（这个方法的好处是保证了分开的两段路径不在同一子树内），但是要使用平衡树，代码复杂度高

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;

const int maxn = 200010;
const int INF = 1000000007;

int N, K; 
int mx, tot, root, ans, tail;

struct Node{
	int u, bel, dis;
	
	bool operator < (const Node &x) const{
		return dis < x.dis;
	} 
}a[maxn]; 

void Add_Node(int u, int bel, int dis){ a[++tail].u = u, a[tail].bel = bel, a[tail].dis = dis; }

int h[maxn], cnt;
struct{
	int to, next;
	int cost;
}e[maxn << 1];
void add(int u, int v, int w){
	e[++cnt].to = v;
	e[cnt].cost = w;
	e[cnt].next = h[u];
	h[u] = cnt;
}

int sz[maxn], son[maxn], dis[maxn], vis[maxn];
int c[maxn]; // c[i] ：属于 i 子树的节点有多少个 

void Get_Root(int u, int par){ // 找重心
	sz[u] = 1, son[u] = 0;
	for(int i = h[u]; i != -1; i = e[i].next){
		int v = e[i].to;
		if(vis[v] || v == par) continue;
		Get_Root(v, u);
		sz[u] += sz[v];
		son[u] = max(son[u], sz[v]);
	} 
	son[u] = max(son[u], tot - son[u]);
	if(mx > son[u]){
		root = u;
		mx = son[u];
	}
}

void Get_Dis(int u, int par, int rt){ // 统计距离
	for(int i = h[u]; i != -1; i = e[i].next){
		int v = e[i].to, w = e[i].cost;
		if(vis[v] || v == par) continue;
		dis[v] = dis[u] + w;
		Add_Node(v, rt, dis[v]);
		Get_Dis(v, u, rt);
	}
}

void calc(int u){
	dis[u] = 0; Add_Node(u, 0, 0);
	for(int i = h[u] ; i != -1; i = e[i].next){
		int v = e[i].to, w = e[i].cost;
		if(vis[v]) continue;
		dis[v] = dis[u] + w;
		Add_Node(v, v, dis[v]);
		Get_Dis(v, u, v);
	} 

	sort(a + 1, a + 1 + tail); // 将子树节点按距离排序
	for(int i = 1; i <= tail ; ++i ) ++c[a[i].bel];  // 统计每个子树内节点的数量

	
	// 统计答案 
	int head;
	for(head = 1; head < tail ; ++head){ 
		--c[a[head].bel];
		while(a[tail].dis + a[head].dis > K){
			--c[a[tail].bel];
			--tail;
			if(tail == head) break; // 防止越界 
		}
		ans += tail - head - c[a[head].bel];
	}
	for(int i = head; i <= tail ; ++i ) --c[a[i].bel];
}

void fenzhi(int u){
	vis[u] = 1;
	tail = 0;
	calc(u);
	
	for(int i = h[u]; i != -1; i = e[i].next){
		int v = e[i].to , w = e[i].cost;
		if(vis[v]) continue;
		
		mx = INF, root = 0, tot = sz[v];
		Get_Root(v,0);

		fenzhi(root);
	}
}

ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }

int main(){
	while(1){
		memset(vis, 0, sizeof(vis));
		memset(h, -1, sizeof(h)); cnt = 0; ans = 0;
		N = read(), K = read();
		if(!N && !K) break;
		
		int u, v, w;
		for(int i = 1; i < N; ++i){
			u = read(), v = read(), w = read();
			++u, ++v;
			add(u, v, w), add(v, u, w);
		}
		
		mx = INF, tot = N;
		Get_Root(1, 0);
		
		fenzhi(root);
		
		printf("%d\n",ans); 
	}
	
	return 0;
}