HDU-2586 (LCA)

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

The Source
ECJTU the Spring Contest The 2009
entitled effect: firstly given a tree, then the distance is determined between two points in the tree.
Problem-solving ideas: This question is a question of LCA template, we first tree in the given point to first find the root of the distance, which is equivalent to a prefix and then we find two points common ancestor, and then from the two root nodes to their common ancestor adding subtracting twice the distance to get to the root of the distance between two nodes pull.
Code:

#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;
        }
    }
}