#10134. 「一本通 4.4 练习 1」Dis

Archie

练习一下树刨

只要记录一下到链顶的距离以及到父亲的距离就行了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m;
struct e{
	int to;
	int ne;
	int k;
}ed[50005];
int x,y,z;
int p;
int head[40005];
int son[40005];
int dep[40005];
int len[40005];
int siz[40005];
int top[40005];
int fa[40005];
int dfn[40005];
int rnk[40005];
int lenf[40005];
void add(int f,int to,int v){
	p++;
	ed[p].k=v;
	ed[p].ne=head[f];
	ed[p].to=to;
	head[f]=p;
}
void dfs(int no,int f){
	son[no]=-1;
	siz[no]=1;
	fa[no]=f;
	dep[no]=dep[f]+1;
	for(int i=head[no];i;i=ed[i].ne){
		if(ed[i].to==f) continue;
		dfs(ed[i].to,no);
		siz[no]+=siz[ed[i].to];
		lenf[ed[i].to]=ed[i].k;
		if(son[no]==-1||siz[ed[i].to]>siz[son[no]]){
			son[no]=ed[i].to;
		}
	}
}
int cnt;
void dfs2(int no,int to,int l){
//	cout<<no<<" "<<to<<endl; 
	cnt++;
	dfn[no]=cnt;
	top[no]=to;
	rnk[cnt]=no;
	if(to!=no)
	{
		len[no]=len[fa[no]]+l;
	}
	if(-1==son[no]) return ;
	dfs2(son[no],to,lenf[son[no]]);
	for(int i=head[no];i;i=ed[i].ne){
		if(ed[i].to==fa[no]||ed[i].to==son[no]) continue;
	//	cout<<"S";
		dfs2(ed[i].to,ed[i].to,0);
	}
	
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<n;++i){
		scanf("%d%d%d",&x,&y,&z);
		add(x,y,z);
		add(y,x,z);
	} 
	dep[1]=1;
	dfs(1,1);
	dfs2(1,1,0);
	while(m--){
		scanf("%d%d",&x,&y);
		int ans=0;
		while(top[x]!=top[y]){
			if(dep[top[x]]>=dep[top[y]]){
				ans+=len[x]+lenf[top[x]]; 
				x=fa[top[x]];
				
			}else{
				ans+=len[y]+lenf[top[y]];
				y=fa[top[y]];
				
			}
		} 
		if(dep[x]>dep[y]){
			
			ans+=len[x]-len[y];
		}else{
			ans+=len[y]-len[x];
		}
		cout<<ans<<endl;;
	}
	
	return 0;
}