LCA进阶
水
1. 欧拉序+ST表
欧拉序求LCA
// Problem: P3379 【模板】最近公共祖先（LCA）
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3379
// Memory Limit: 512 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#include <bits/extc++.h>
#define INF 0x7fffffff
#define MAXN 500002
#define MAXM 10003
#define eps 1e-9
#define foru(a,b,c) for(int a=b;a<=c;a++)
#define ford(a,b,c) for(int a=b;a>=c;a--)
#define RT return 0;
#define db(x) cout<<endl<<x<<endl;
#define LL long long
#define LXF int
#define RIN rin()
#define HH printf("\n")
using namespace std;
inline LXF rin(){
LXF x=0,w=1;
char ch=0;
while(ch<'0'||ch>'9'){
if(ch=='-') w=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9'){
x=x*10+(ch-'0');
ch=getchar();
}
return x*w;
}
inline void out(LXF x){
if(x<0){
x=-x;
putchar('-');
}
if(x>9) out(x/10);
putchar(x%10+'0');
}
int n,m,s,cnt,to[MAXN];
struct N{
int d,id;
}st[22][MAXN<<2];
bool operator < (const N &p,const N &q){
return p.d<q.d;
}
vector<int> e[MAXN];
void dfs(int s,int fa,int dep){
to[s]=++cnt;
st[0][cnt]=(N){dep,s};
for(int i=0;i<e[s].size();i++){
int v=e[s][i];
if(v!=fa){
dfs(v,s,dep+1);
st[0][++cnt]=(N){dep,s};
}
}
}
int ask(int l,int r){
int k=log2(r-l+1);
return min(st[k][l],st[k][r-(1<<k)+1]).id;
}
int main(){
n=RIN,m=RIN,s=RIN;
foru(i,1,n-1){
int u=RIN,v=RIN;
e[u].push_back(v);
e[v].push_back(u);
}
dfs(s,-1,1);
for(int i=1;i<=21;i++){
for(int j=1;j+(1<<i)-1<=cnt;j++){
st[i][j]=min(st[i-1][j],st[i-1][j+(1<<(i-1))]);
}
}
foru(i,1,m){
int l=RIN,r=RIN;
printf("%d\n",ask(min(to[l],to[r]),max(to[l],to[r])));
}
return 0;
}2. 树链剖分求LCA
树剖求LCA
// Problem: P3379 【模板】最近公共祖先（LCA）
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3379
// Memory Limit: 512 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#include <bits/extc++.h>
#define INF 0x7fffffff
#define MAXN 500005
#define MAXM 500005
#define eps 1e-9
#define foru(a,b,c) for(int a=b;a<=c;a++)
#define ford(a,b,c) for(int a=b;a>=c;a--)
#define RT return 0;
#define db(x) cout<<endl<<x<<endl;
#define LL long long
#define LXF int
#define RIN rin()
#define HH printf("\n")
using namespace std;
inline LXF rin(){
LXF x=0,w=1;
char ch=0;
while(ch<'0'||ch>'9'){
if(ch=='-') w=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9'){
x=x*10+(ch-'0');
ch=getchar();
}
return x*w;
}
inline void out(LXF x){
if(x<0){
x=-x;
putchar('-');
}
if(x>9) out(x/10);
putchar(x%10+'0');
}
int n,m,r,top[MAXN],dep[MAXN],cnt,fa[MAXN],siz[MAXN],son[MAXN],id[MAXN];
vector<int> e[MAXN];
void dfs1(int s,int fath){
dep[s]=dep[fath]+1;
fa[s]=fath;
siz[s]=1;
int maxson=-1;
for(int i=0;i<e[s].size();i++){
int v=e[s][i];
if(v!=fath){
dfs1(v,s);
siz[s]+=siz[v];
if(siz[v]>maxson){
son[s]=v;
maxson=siz[v];
}
}
}
}
void dfs2(int s,int topf){
id[s]=++cnt;
top[s]=topf;
if(!son[s]) return ;
dfs2(son[s],topf);
for(int i=0;i<e[s].size();i++){
int v=e[s][i];
if(v!=fa[s]&&v!=son[s]){
dfs2(v,v);
}
}
}
int LCA(int x,int y){
while(top[x]!=top[y]){
if(dep[top[x]]<dep[top[y]]) swap(x,y);
x=fa[top[x]];
}
if(dep[x]<dep[y]) swap(x,y);
return y;
}
int main(){
n=RIN,m=RIN,r=RIN;
foru(i,1,n-1){
int u=RIN,v=RIN;
e[u].push_back(v);
e[v].push_back(u);
}
dfs1(r,0);
dfs2(r,r);
while(m--){
printf("%d\n",LCA(RIN,RIN));
}
return 0;
}3. (离线) Tarjan算法求LCA
记得用前向星存树，vector会慢1.5s以上
Tarjan求LCA
// Problem: P3379 【模板】最近公共祖先（LCA）
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3379
// Memory Limit: 512 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#include <bits/extc++.h>
#define INF 0x7fffffff
#define MAXN 500005
#define MAXM 500005
#define eps 1e-9
#define foru(a,b,c) for(int a=b;a<=c;a++)
#define ford(a,b,c) for(int a=b;a>=c;a--)
#define RT return 0;
#define db(x) cout<<endl<<x<<endl;
#define LL long long
#define LXF int
#define RIN rin()
#define HH printf("\n")
using namespace std;
inline LXF rin(){
LXF x=0,w=1;
char ch=0;
while(ch<'0'||ch>'9'){
if(ch=='-') w=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9'){
x=x*10+(ch-'0');
ch=getchar();
}
return x*w;
}
inline void out(LXF x){
if(x<0){
x=-x;
putchar('-');
}
if(x>9) out(x/10);
putchar(x%10+'0');
}
int n,m,r,jh[MAXN],ehead[MAXN],qhead[MAXN],ecnt,qcnt;
bool vis[MAXN];
int find(int x){
return jh[x]==0?x:jh[x]=find(jh[x]);
}
struct E{
int v,nxt,lca;
}e[MAXM<<1],q[MAXM<<1];
void tarjan(int s){
vis[s]=1;
for(int i=ehead[s];i;i=e[i].nxt){
int v=e[i].v;
if(!vis[v]){
tarjan(v);
jh[v]=find(s);
}
}
for(int i=qhead[s];i;i=q[i].nxt){
int v=q[i].v;
if(vis[v]){
if(i&1) q[i].lca=q[i+1].lca=find(v);
else q[i].lca=q[i-1].lca=find(v);
}
}
}
void add_e(int u,int v){
e[++ecnt]=(E){v,ehead[u],0};
ehead[u]=ecnt;
}
void add_q(int u,int v){
q[++qcnt]=(E){v,qhead[u],0};
qhead[u]=qcnt;
}
int main(){
n=RIN,m=RIN,r=RIN;
foru(i,1,n-1){
int u=RIN,v=RIN;
add_e(u,v);
add_e(v,u);
}
foru(i,1,m){
int u=RIN,v=RIN;
add_q(u,v);
add_q(v,u);
}
tarjan(r);
foru(i,1,m){
printf("%d\n",q[i<<1].lca);
}
return 0;
}总结
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