LCA
hdu2586，生成树两点间的距离
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
题目链接
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;
}
