具体讲解可看：https://www.cnblogs.com/zhouzhendong/p/7256007.html

LCA_Tarjan

Tarjan 算法求 LCA 的时间复杂度为 O((n+q)α(n)) ，是一种离线算法，要用到并查集。

#include <bits/stdc++.h>
using namespace std;
const int N=40000+5;
struct Edge{
    int cnt,x[N],y[N],z[N],nxt[N],fst[N];
    void set(){
        cnt=0;
        memset(x,0,sizeof x);
        memset(y,0,sizeof y);
        memset(z,0,sizeof z);
        memset(nxt,0,sizeof nxt);
        memset(fst,0,sizeof fst);
    }
    void add(int a,int b,int c){
        x[++cnt]=a;
        y[cnt]=b;
        z[cnt]=c;
        nxt[cnt]=fst[a];
        fst[a]=cnt;
    }
}e,q;
int T,n,m,from,to,dist,in[N],rt,dis[N],fa[N],ans[N];
bool vis[N];
void dfs(int rt){
    for (int i=e.fst[rt];i;i=e.nxt[i]){
        dis[e.y[i]]=dis[rt]+e.z[i];
        dfs(e.y[i]);
    }
}
int getf(int k){
    return fa[k]==k?k:fa[k]=getf(fa[k]);
}
void LCA(int rt){
    for (int i=e.fst[rt];i;i=e.nxt[i]){
        LCA(e.y[i]);
        fa[getf(e.y[i])]=rt;
    }
    vis[rt]=1;
    for (int i=q.fst[rt];i;i=q.nxt[i])
        if (vis[q.y[i]]&&!ans[q.z[i]])
            ans[q.z[i]]=dis[q.y[i]]+dis[rt]-2*dis[getf(q.y[i])];
}
int main(){
    scanf("%d",&T);
    while (T--){
        q.set(),e.set();
        memset(in,0,sizeof in);
        memset(vis,0,sizeof vis);
        memset(ans,0,sizeof ans);
        scanf("%d%d",&n,&m);
        for (int i=1;i<n;i++)
            scanf("%d%d%d",&from,&to,&dist),e.add(from,to,dist),in[to]++;
        for (int i=1;i<=m;i++)
            scanf("%d%d",&from,&to),q.add(from,to,i),q.add(to,from,i);
        rt=0;
        for (int i=1;i<=n&&rt==0;i++)
            if (in[i]==0)
                rt=i;
        dis[rt]=0;
        dfs(rt);
        for (int i=1;i<=n;i++)
            fa[i]=i;
        LCA(rt);
        for (int i=1;i<=m;i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}

倍增

我们可以用倍增来在线求 LCA ，时间和空间复杂度分别是 O((n+q)logn)和 O(nlogn) 

#include <bits/stdc++.h>
using namespace std;
const int N=10000+5;
vector <int> son[N];
int T,n,depth[N],fa[N],in[N],a,b;
void dfs(int prev,int rt){
    depth[rt]=depth[prev]+1;
    fa[rt]=prev;
    for (int i=0;i<son[rt].size();i++)
        dfs(rt,son[rt][i]);
}
int LCA(int a,int b){
    if (depth[a]>depth[b])
        swap(a,b);
    while (depth[b]>depth[a])
        b=fa[b];
    while (a!=b)
        a=fa[a],b=fa[b];
    return a;
}
int main(){
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
        for (int i=1;i<=n;i++)
            son[i].clear();
        memset(in,0,sizeof in);
        for (int i=1;i<n;i++){
            scanf("%d%d",&a,&b);
            son[a].push_back(b);
            in[b]++;
        }
        depth[0]=-1;
        int rt=0;
        for (int i=1;i<=n&&rt==0;i++)
            if (in[i]==0)
                rt=i;
        dfs(0,rt);
        scanf("%d%d",&a,&b);
        printf("%d\n",LCA(a,b));
    }
    return 0;
}

RMQ

RMQ可以 O(nlogn)预处理，O(1)在线查询的算法

#include <bits/stdc++.h>
#define time _____time
using namespace std;
const int N=50005;
struct Gragh{
    int cnt,y[N*2],z[N*2],nxt[N*2],fst[N];
    void clear(){
        cnt=0;
        memset(fst,0,sizeof fst);
    }
    void add(int a,int b,int c){
        y[++cnt]=b,z[cnt]=c,nxt[cnt]=fst[a],fst[a]=cnt;
    }
}g;
int n,m,depth[N],in[N],out[N],time;
int ST[N*2][20];
void dfs(int x,int pre){
    in[x]=++time;
    ST[time][0]=x;
    for (int i=g.fst[x];i;i=g.nxt[i])
        if (g.y[i]!=pre){
            depth[g.y[i]]=depth[x]+g.z[i];
            dfs(g.y[i],x);
            ST[++time][0]=x;
        }
    out[x]=time;
}
void Get_ST(int n){
    for (int i=1;i<=n;i++)
        for (int j=1;j<20;j++){
            ST[i][j]=ST[i][j-1];
            int v=i-(1<<(j-1));
            if (v>0&&depth[ST[v][j-1]]<depth[ST[i][j]])
                ST[i][j]=ST[v][j-1];
        }
}
int RMQ(int L,int R){
    int val=floor(log(R-L+1)/log(2));
    int x=ST[L+(1<<val)-1][val],y=ST[R][val];
    if (depth[x]<depth[y])
        return x;
    else
        return y;
}
int main(){
    scanf("%d",&n);
    for (int i=1,a,b,c;i<n;i++){
        scanf("%d%d%d",&a,&b,&c);
        a++,b++;
        g.add(a,b,c);
        g.add(b,a,c);
    }
    time=0;
    dfs(1,0);
    depth[0]=1000000;
    Get_ST(time);
    scanf("%d",&m);
    while (m--){
        int x,y;
        scanf("%d%d",&x,&y);
        if (in[x+1]>in[y+1])
            swap(x,y);
        int LCA=RMQ(in[x+1],in[y+1]);
        printf("%d\n",depth[x+1]+depth[y+1]-depth[LCA]*2);
    }
    return 0;
}