HDOJ 2586 How far away ？

tarjanLCA求树上任意两点间的距离。。。。

任意选一个点当根节点，dist[a,b]=   dist[a] +dist[b] -2* lca[a,b]

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4254    Accepted Submission(s): 1622

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

 

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#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn=44000;

struct Edge
{
	int to,next,distan;
}edge[2*maxn],edge2[maxn];

int Adj[maxn],Adj2[maxn],Size,Size2;

void init_1()
{
	Size=0;
	memset(Adj,-1,sizeof(Adj));
}

void init_2()
{
	Size2=0;
	memset(Adj2,-1,sizeof(Adj2));
}

void Add_Edge(int u,int v,int c)
{
	edge[Size].to=v;
	edge[Size].next=Adj[u];
	edge[Size].distan=c;
	Adj[u]=Size++;
}

void Add_Edge2(int u,int v,int id)
{
	edge2[Size2].to=v;
	edge2[Size2].next=Adj2[u];
	edge2[Size2].distan=id;
	Adj2[u]=Size2++;
}

int n,m;

int fa[maxn];
int Find(int x)
{
	if(x==fa[x]) return x;
	return fa[x]=Find(fa[x]);
}

bool vis[maxn];
int Distan[maxn],ans[maxn];

void LCA(int u,int father)
{
	for(int i=Adj[u];~i;i=edge[i].next)
	{
		int v=edge[i].to;
		if(v==father) continue;
		Distan[v]=Distan[u]+edge[i].distan;
		LCA(v,u);
		fa[v]=u;		
	}
	vis[u]=true;
	for(int i=Adj2[u];~i;i=edge2[i].next)
	{
		int v=edge2[i].to;
		int id=edge2[i].distan;
		if(vis[v])	
		{
			ans[id]=Distan[u]+Distan[v]-2*Distan[Find(v)];
		}
	}
}

int main()
{
	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		init_1();init_2();
		scanf("%d%d",&n,&m);
		fa[0]=0;fa[1]=1;fa[2]=2;
		for(int i=0;i<n-1;i++)
		{
 			int a,b,c;
 			scanf("%d%d%d",&a,&b,&c);
 			Add_Edge(a,b,c);
 			Add_Edge(b,a,c);
 			fa[i+3]=i+3;
		}
		for(int i=0;i<m;i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			Add_Edge2(a,b,i);
			Add_Edge2(b,a,i);
		}

		memset(Distan,0,sizeof(Distan));
		memset(ans,0,sizeof(ans));
		memset(vis,false,sizeof(vis));
		LCA(1,1);
		for(int i=0;i<m;i++)
			printf("%d\n",ans[i]);
	}
	return 0;
}