LCA

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    
Memory Limit: 32768/32768 K (Java/Others)

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 $T(T \leqslant 10)$, indicating the number of test cases.
For each test case,in the first line there are two numbers $n(2 \leqslant n\leqslant 40000)$ and $m (1 \leqslant m \leqslant 200)$,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 $k(0<k \leqslant 40000)$.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

题目链接

hdu2586

题目描述

给定一个生成树，求任意两个结点的距离

解题思路

1.树上倍增法

• 时间复杂度：$((n+m)log(n))$

2.tarjan优化
本质：一种离线做法，并查集对“向上标记法”的优化

• 时间复杂度：$o(n+m)$

代码

• 常规代码

#include<bits/stdc++.h>
using namespace std;
const int N=40010;
vector<pair<int,int>>adj[N];//邻接表
int dist[N],d[N],f[N][20];//dist[i]:结点i到根节点的距离，d[i]:结点i的深度，
                            //f[i][j]:结点i向上走2^j步到达的结点
int T,n,m,x,y,z,t;
void bfs()
{
    queue<int>q;
    q.push(1);//设1为根节点
    dist[1]=0;
    d[1]=1;
    while(q.size())
    {
        int u=q.front();
        q.pop();
        for(auto p:adj[u])
        {
            int v=p.first,w=p.second;
            if(dist[v]||v==1)continue;//注意v==1的情况
            d[v]=d[u]+1;
            dist[v]=dist[u]+w;
            f[v][0]=u;
            for(int i=1;i<=t;i++)f[v][i]=f[f[v][i-1]][i-1];//注意，这里是递推式子，应该从小到大
            q.push(v);
        }
    }
}
int lca(int x,int y)
{
    if(d[x]>d[y])swap(x,y);
    for(int i=t;i>=0;i--)
        if(d[f[y][i]]>=d[x])y=f[y][i];
    if(x==y)return x;
    for(int i=t;i>=0;i--)
        if(f[x][i]!=f[y][i])x=f[x][i],y=f[y][i];
    return f[x][0];
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        t=log(n)/log(2)+1;
        for(int i=1;i<n;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            adj[x].emplace_back(y,z);
            adj[y].emplace_back(x,z);
        }
        bfs();//预处理
        while(m--)
        {
            scanf("%d%d",&x,&y);
            printf("%d\n",dist[x]+dist[y]-2*dist[lca(x,y)]);
        }
        for(int i=1;i<=n;i++)d[i]=0,dist[i]=0,adj[i].clear();//多组测试，清零
    }
    return 0;
}

• 高效代码

#include<bits/stdc++.h>
using namespace std;
const int N=40010;
vector<int>query[N],query_id[N];
vector<pair<int,int>>adj[N];
int T,n,m,x,y,z,fa[N],dist[N],ans[210],vis[N];//ans记录答案,v标记是否遍历
void add_query(int u,int v,int id)//query与query_id同步，离线记录
{
    query[u].emplace_back(v),query_id[u].emplace_back(id);
    query[v].emplace_back(u),query_id[v].emplace_back(id);
}
int get(int x)
{
    if(x==fa[x])return x;
    return fa[x]=get(fa[x]);
}
void tarjan(int u)
{
    vis[u]=1;
    for(auto p:adj[u])
    {
        int v=p.first,w=p.second;
        if(vis[v])continue;
        dist[v]=dist[u]+w;
        tarjan(v);
        fa[v]=u;
    }
    for(int i=0;i<query[u].size();i++)
    {
        int v=query[u][i],id=query_id[u][i];
        if(vis[v]==2)//已回溯的点才有更新过的father
        {
            int lca=get(v);
            ans[id]=dist[u]+dist[v]-2*dist[lca];
        }
    }
    vis[u]=2;
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            fa[i]=i;
            dist[i]=vis[i]=0;
            adj[i].clear();
            query[i].clear();
            query_id[i].clear();
        }
        for(int i=1;i<n;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            adj[x].emplace_back(y,z);
            adj[y].emplace_back(x,z);
        }
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            if(x==y)ans[i]=0;
            else
                add_query(x,y,i);
        }
        tarjan(1);
        for(int i=1;i<=m;i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}

__EOF__

本文作者：acwing_zyy
本文链接：https://www.cnblogs.com/zyyun/p/15139481.html
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