hdu:How far away ？（树链剖分）

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

数据结构-LCA(树链剖分)

用树链剖分的方式求LCA,用edg数组保留点到父结点的距离,dis数组保存点到top结点的距离,在求LCA的过程中求出答案

点击查看代码

#include<bits/stdc++.h>
using namespace std;
const int N=1e5;
int h[N],e[N],ne[N],c[N],idx;
int s[N],d[N],f[N],son[N],top[N],dis[N],edg[N];
void add(int u,int v,int w)
{
    e[++idx]=v;
    ne[idx]=h[u];
    c[idx]=w;
    h[u]=idx;
}
void dfs1(int x,int fa,int w)
{
    s[x]=1;d[x]=d[fa]+1;
    son[x]=0;f[x]=fa;edg[x]=w;
    for(int i=h[x];~i;i=ne[i])
    {
        int j=e[i];
        if(j==fa) continue;
        dfs1(j,x,c[i]);
        s[x]+=s[j];
        if(s[son[x]]<s[j]) son[x]=j;
    }
}
void dfs2(int x,int t)
{
    top[x]=t;
    if(x!=t)
    dis[x]=dis[f[x]]+edg[x];
    if(son[x]!=0) dfs2(son[x],t);
    for(int i=h[x];~i;i=ne[i])
    {
        int j=e[i];
        if(j!=f[x]&&j!=son[x]) dfs2(j,j);
    }
}
int lca(int u,int v)
{
    int ans=0;
    while(top[u]!=top[v])
    {
        if(d[u]<d[v]) swap(u,v);
        ans+=dis[u]+edg[top[u]];
        u=f[top[u]];
    }
    ans+=abs(dis[u]-dis[v]);
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int T;
    cin>>T;
    while(T--)
    {
        memset(h,-1,sizeof h);
        idx=0;
        int n,m;
        cin>>n>>m;
        for(int i=0;i<n-1;++i)
        {
            int u,v,w;
            cin>>u>>v>>w;
            add(u,v,w);
            add(v,u,w);
        }
        dfs1(1,0,0);
        dfs2(1,1);
        for(int i=0;i<m;++i)
        {
            int a,b;
            cin>>a>>b;
            cout<<lca(a,b)<<'\n';
        }
        
    }
}