给一棵树，每条边有权。求一条简单路径，权值和等于 (k)，且边的数量最小。

分析

点分治，记录一定权值的最小边数量，
每遍历一棵子树后统计答案

代码

#include <cstdio>
#include <cctype>
#include <algorithm>
#define rr register
using namespace std;
const int N=200011; bool v[N]; struct node{int y,w,next;}e[N<<1];
int big[N],siz[N],SIZ,root,tot,ans,w[N*5],dep[N],dis[N],n,m,as[N],et=1;
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans; 
}
inline void Max(int &a,int b){a=a>b?a:b;}
inline void Min(int &a,int b){a=a<b?a:b;}
inline void dfs(int x,int fa){
	siz[x]=1,big[x]=0;
	for (rr int i=as[x];i;i=e[i].next)
	if (e[i].y!=fa&&!v[e[i].y]){
		dfs(e[i].y,x);
		siz[x]+=siz[e[i].y];
		Max(big[x],siz[e[i].y]);
	}
	Max(big[x],SIZ-siz[x]);
	if (big[x]<big[root]) root=x;
}
inline void Get(int x,int fa,int Dis,int Dep){
	if (Dis>m) return;
	dis[++tot]=Dis,dep[tot]=Dep;
	for (rr int i=as[x];i;i=e[i].next)
	if (e[i].y!=fa&&!v[e[i].y])
	    Get(e[i].y,x,Dis+e[i].w,Dep+1);
}
inline void calc(int x){
	w[0]=tot=0;
	for (rr int i=as[x];i;i=e[i].next)
	if (!v[e[i].y]){
		rr int Tot=tot;
		Get(e[i].y,x,e[i].w,1);
		for (rr int j=Tot+1;j<=tot;++j) Min(ans,w[m-dis[j]]+dep[j]);
		for (rr int j=Tot+1;j<=tot;++j) Min(w[dis[j]],dep[j]);
	}
	for (rr int i=1;i<=tot;++i) w[dis[i]]=w[m+1];
}
inline void dp(int x){
	v[x]=1,calc(x);
	for (rr int i=as[x];i;i=e[i].next)
	if (!v[e[i].y]){
		big[0]=SIZ=siz[e[i].y];
		dfs(e[i].y,root=0),dp(root);
	}
}
signed main(){
	n=iut(),m=iut(),ans=n+1;
	for (rr int i=1;i<=m+1;++i) w[i]=ans;
	for (rr int i=1;i<n;++i){
		rr int x=iut()+1,y=iut()+1,w=iut();
		e[++et]=(node){y,w,as[x]},as[x]=et;
		e[++et]=(node){x,w,as[y]},as[y]=et;
	}
	big[0]=SIZ=n,dfs(1,root=0),dp(root);
	if (ans>n) printf("-1");
	    else printf("%d",ans);
	return 0;
}