大意

给定一棵树，求出一条路径$p$p，使得其的权值$S\leq k\leq E$S≤k≤E

思路

正解点分治

$dfs$dfs代码(70)

#include<cstdio>
#include<algorithm>
#define ri register int
using namespace std;int n,S,E,l[100001],x,y,w,tot,minn=2147483647;
struct node{int next,to,w;}e[200001];
inline void add(ri u,ri v,ri w){e[++tot]=(node){l[u],v,w};l[u]=tot;return;}
bool vis[100001];
inline void dfs(ri x,ri now,ri from)
{
	if(now>=S)
	{
		if(now>=minn) return;
		minn=now;
	}
	if(now>E) return;
	for(ri i=l[x];i;i=e[i].next)
	{
		int y=e[i].to,w=e[i].w;
		if(y!=from)	dfs(y,now+w,x);
	}
	return;
}
signed main()
{
	scanf("%d%d%d",&n,&S,&E);
	for(ri i=1;i<n;i++) 
	{
		scanf("%d%d%d",&x,&y,&w);
		if(w<=E)
		{
			add(x,y,w);add(y,x,w);
			if(w>=S&&w<minn) minn=w;
		}
	}
	if(minn<2147483647) return printf("%d",minn)&0;
	for(ri i=1;i<=n;i++) dfs(i,0,-1);
	if(minn<2147483647) return printf("%d",minn)&0;
	puts("-1");
}

点分治代码

#include<cstdio>
#include<algorithm>
#define ri register int
using namespace std;int n,S,E,l[100010],x,y,w,tot,Ans=2147483647;
struct node{int next,to,w;}e[200020];
struct Node{int len,father;}s[100010];
int sz[100010],rt,f[100010]={1<<27},d[100010],t,fs[100010],nt[100010],ns;
bool vis[100010];
inline void add(ri u,ri v,ri w){e[++tot]=(node){l[u],v,w};l[u]=tot;return;}
inline void gmin(ri &x,ri y){if(x>y)x=y;return;}
inline void gmax(ri &x,ri y){if(x<y)x=y;return;}
inline bool cmp(Node x,Node y){return x.len<y.len;}
inline void groot(ri x,ri fa)
{
	sz[x]=1;f[x]=0;
	for(ri i=l[x];i;i=e[i].next)
	{
		int y=e[i].to;
		if(y==fa||vis[y]) continue;
		groot(y,x);
		sz[x]+=sz[y];gmax(f[x],sz[y]);
	}
	gmax(f[x],ns-sz[x]);
	if(f[x]<f[rt]) rt=x;
	return;
}
inline void gparent(ri x,ri fa)
{
	s[++t]=(Node){d[x],fs[x]};
	for(ri i=l[x];i;i=e[i].next)
	{
		int y=e[i].to,w=e[i].w;
		if(y==fa||vis[y]) continue;
		if(fa)fs[y]=fs[x];d[y]=d[x]+w;
		gparent(y,x);
	}
	return;
}
inline void gcal(ri x)
{
	d[x]=0;t=0;fs[x]=x;
	for(ri i=l[x];i;i=e[i].next) fs[e[i].to]=e[i].to;
	gparent(x,0);sort(s+1,s+1+t,cmp);nt[t]=t+1;
	for(ri i=t-1;i;i--) nt[i]=s[i].father==s[i+1].father?nt[i+1]:i+1;
	for(ri l=1,r=t;l<r;++l)
	{
		for(;l<r&&s[r-1].len+s[l].len>=S;r--);
		if(s[l].father==s[r].father) r=nt[r];
		if(r<=t&&s[l].len+s[r].len>=S) gmin(Ans,s[l].len+s[r].len);
	}
	return;
}
inline void dfs(ri x)
{
	gcal(x);vis[x]=true;
	for(ri i=l[x];i;i=e[i].next)
	{
		int y=e[i].to;
		if(!vis[y])
		{
			rt=0;ns=sz[x];
			groot(y,x);dfs(rt);
		}
	}
	return;
}
signed main()
{
	scanf("%d%d%d",&n,&S,&E);
	for(ri i=1;i<n;i++) 
	{
		scanf("%d%d%d",&x,&y,&w);
		add(x,y,w);add(y,x,w);
	}
	ns=n;groot(1,0);
	dfs(rt);
	if(Ans>E) puts("-1");else printf("%d",Ans);
}