How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32298 Accepted Submission(s): 13051
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
题目大意为：先给出一棵树，然后求出这棵树中两个点之间的距离。
解题思路：这道题是一个LCA的模板题，我们先在给出的树中先求出该点到根节点的距离，这就相当于一个前缀和，然后我们在求出两个点的最近公共祖先，然后再将两个结点到根节点的距离相加在减去他们公共祖先到根节点的距离的2倍就得到了这两个结点之间的距离拉。
代码：
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e6+7;
const int inf=0x3f3f3f3f;
#define int long long
struct edge
{
int v,w,next;
}e[maxn];
int cnt,head[maxn];
void add(int a,int b,int c)
{
e[++cnt]=edge{b,c,head[a]};
head[a]=cnt;
}
int dis[maxn],fa[maxn][20],lg[maxn],depth[maxn],n,m;
void init()
{
for(int i=1;i<40010;i++){
lg[i]=lg[i-1]+((1<<lg[i-1])==i);
}
}
void dfs(int u,int f)
{
depth[u]=depth[f]+1;
fa[u][0]=f;
for(int i=1;i<lg[depth[u]];i++){
fa[u][i]=fa[fa[u][i-1]][i-1];
}
for(int i=head[u];i;i=e[i].next){
int v=e[i].v,w=e[i].w;
if(v==f)continue;
dis[v]=dis[u]+w;
dfs(v,u);
}
}
int lca(int a,int b)
{
if(depth[a]<depth[b]){
swap(a,b);
}
while(depth[a]>depth[b]){
int k=lg[depth[a]-depth[b]]-1;
a=fa[a][k];
}
if(a==b)return a;
for(int i=lg[depth[a]]-1;i>=0;i--){
if(fa[a][i]!=fa[b][i]){
a=fa[a][i],b=fa[b][i];
}
}
return fa[a][0];
}
void debug()
{
for(int i=1;i<=n;i++){
for(int j=0;j<lg[depth[i]];j++)cout<<fa[i][j]<<' ';cout<<endl;
}
}
signed main()
{
int t;
cin>>t;
init();
while(t--){
//int n,m;
cin>>n>>m;
//scanf("%d%d",&n,&m);
memset(head,0,sizeof head);
//memset(dis,0,sizeof dis);
cnt=0;
for(int i=1;i<n;i++){
int u,v,w;
cin>>u>>v>>w;
//scanf("%d%d%d",&u,&v,&w);
add(u,v,w);add(v,u,w);
}
dfs(1,0);
//debug();
while(m--){
int a,b;
cin>>a>>b;
//scanf("%d%d",&a,&b);
int x=lca(a,b);
//cout<<"lca="<<x<<endl;
cout<<dis[a]+dis[b]-2*dis[x]<<endl;
}
}
}
