hdu 2586

How far away ？

Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

 

Input

First line is a single integer T(T<=10), indicating the number of test cases.
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

 

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

 

Sample Input

2 3 2 1 2 10 3 1 15 1 2 2 3 2 2 1 2 100 1 2 2 1

 

Sample Output

10 25 100 100

 

Source

ECJTU 2009 Spring Contest

 

Recommend

lcy

题意：给一个无根树，有q个询问，问两点之间的距离。

思路：LCA模板题

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#define M 100100
using namespace std;
struct node{int w,next,y;}a[4*M];
struct tree{int par[30],dis,dep;}t[4*M];
int head[M*4];
int m,n,l,y,x,k;
char s[100];
void add(int x,int y,int k)
{
	a[++l].y=y;a[l].next=head[x];a[l].w=k;head[x]=l;
}
void bt(int x,int fa,int dis)
{   //printf("%d %d %d\n",x,fa,dis);
	t[x].par[0]=fa;t[x].dis=dis;t[x].dep=t[fa].dep+1;
	for(int i=1;(1<<i)<=t[x].dep;i++) t[x].par[i]=t[t[x].par[i-1]].par[i-1];
	for(int k=head[x];k;k=a[k].next)
	   {
	   	int y=a[k].y;
	   	if(!(y==fa)) bt(y,x,dis+a[k].w);
	   }
}
int lca(int x,int y)
{   int tt;
	if(t[x].dep<t[y].dep) {tt=x;x=y;y=tt;}
	for(int i=15;i>=0;i--)
	   if (t[x].dep-t[y].dep>=(1<<i)) x=t[x].par[i];
	if(x==y) return x;
	for(int i=15;i>=0;i--)
	   if(!(t[x].par[i]==t[y].par[i])&&(t[x].dep>=(1<<i)))
	      {
	      	x=t[x].par[i];
	      	y=t[y].par[i];
		  }
	return(t[x].par[0]);
}
void solve()
{
	t[0].dep=-1;
	bt(1,0,0);
}
int main()
{   int T;
    scanf("%d",&T);
	while(T--)
	  {
	  scanf("%d%d",&n,&m);
	  //printf("%d%d",n,m);
	  memset(t[x].par,0,sizeof(t[x].par));l=0;
	  memset(head,0,sizeof(head));
	  for(int i=1;i<n;i++)
	     {
	   	 scanf("%d%d%d",&x,&y,&k);
	   	 add(x,y,k);add(y,x,k);
	     }
		solve();
		for(int i=1;i<=m;i++)
		  {
		  	scanf("%d%d",&x,&y);
		  	//printf("%d %d\n",t[x].dis,t[y].dis);
		  	int k=lca(x,y);
		  	printf("%d\n",t[x].dis+t[y].dis-t[k].dis*2);
		  }
	  gets(s);
      }
}