求树上两条路径的 LCP （树上每个节点代表一个字符）
总共写+调了6个多小时，终于过了~
绝对是我写过的最复杂的数据结构了
我们对这棵树进行轻重链剖分，然后把所有的重链分正串，反串插入到广义后缀自动机中.
求 LCP 的话就是后缀树中两点 $LCA$ 的深度.
如果 $LCP$ 的长度小于两个重链的长度，就直接输出答案，否则还要继续爬重链.
这道题恶心之处在于将所有重链不重不漏存起来，无法形容有多恶心.
code:
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
#include <stack>
#include <string>
using namespace std;
void setIO(string s)
{
string in=s+".in";
string out=s+".out";
freopen(in.c_str(),"r",stdin);
freopen(out.c_str(),"w",stdout);
}
namespace sam
{
#define N 1200004
int last,tot;
int ch[N][27],pre[N],mx[N];
inline int extend(int c)
{
if(ch[last][c])
{
int p=last;
int q=ch[p][c];
if(mx[q]==mx[p]+1) last=q;
else
{
int nq=++tot;
mx[nq]=mx[p]+1;
memcpy(ch[nq],ch[q],sizeof(ch[q]));
pre[nq]=pre[q],pre[q]=nq;
for(;p&&ch[p][c]==q;p=pre[p]) ch[p][c]=nq;
last=nq;
}
}
else
{
int np=++tot,p=last;
mx[np]=mx[p]+1,last=np;
for(;p&&!ch[p][c];p=pre[p]) ch[p][c]=np;
if(!p) pre[np]=1;
else
{
int q=ch[p][c];
if(mx[q]==mx[p]+1) pre[np]=q;
else
{
int nq=++tot;
mx[nq]=mx[p]+1;
memcpy(ch[nq],ch[q],sizeof(ch[q]));
pre[nq]=pre[q],pre[np]=pre[q]=nq;
for(;p&&ch[p][c]==q;p=pre[p]) ch[p][c]=nq;
}
}
last=np;
}
return last;
}
#undef N
};
// 优美的后缀自动机
namespace suffix_tree
{
#define LOG 21
#define N 1200006
vector<int>G[N];
int fa[LOG][N],dis[N];
void dfs(int u)
{
int i,j;
dis[u]=dis[sam::pre[u]]+1;
fa[0][u]=sam::pre[u];
for(i=1;i<LOG;++i) fa[i][u]=fa[i-1][fa[i-1][u]];
for(i=0;i<(int)G[u].size();++i) dfs(G[u][i]);
}
inline void build()
{
int i,j;
for(i=2;i<=sam::tot;++i)
G[sam::pre[i]].push_back(i);
dfs(1);
}
inline int jump(int x,int len)
{
for(int i=LOG-1;i>=0;--i) if(sam::mx[fa[i][x]]>=len) x=fa[i][x];
return x;
}
inline int LCA(int x,int y)
{
if(dis[x]!=dis[y])
{
if(dis[x]>dis[y]) swap(x,y);
for(int i=LOG-1;i>=0;--i) if(dis[fa[i][y]]>=dis[x]) y=fa[i][y];
}
if(x==y) return x;
for(int i=LOG-1;i>=0;--i) if(fa[i][x]!=fa[i][y]) x=fa[i][x], y=fa[i][y];
return fa[0][x];
}
#undef N
#undef LOG
};
// 优美的后缀树
namespace tree
{
#define N 600003
int tot,iii;
int son[N],size[N],dep[N],dfn[N],top[N],fa[N],up[N],down[N],val[N],idx[N],pp[N];
vector<int>G[N];
struct node
{
int u,c;
node(int u=0,int c=0):u(u),c(c){}
};
vector<node>A[N];
inline void add(int u,int v)
{
G[u].push_back(v);
G[v].push_back(u);
}
void dfs1(int u,int ff)
{
fa[u]=ff,dep[u]=dep[ff]+1,size[u]=1;
for(int i=0;i<G[u].size();++i)
{
if(G[u][i]==ff) continue;
dfs1(G[u][i],u);
size[u]+=size[G[u][i]];
if(size[G[u][i]]>size[son[u]]) son[u]=G[u][i];
}
}
void dfs2(int u,int tp)
{
idx[u]=++iii;
pp[iii]=u;
top[u]=tp;
if(tp==u) dfn[tp]=++tot;
A[dfn[tp]].push_back(node(u,val[u]));
if(son[u]) dfs2(son[u],tp);
for(int i=0;i<G[u].size();++i)
if(G[u][i]!=son[u]&&G[u][i]!=fa[u])
dfs2(G[u][i], G[u][i]);
}
inline void build()
{
int i,j;
dfs1(1,0);
dfs2(1,1);
for(i=1;i<=tot;++i)
{
sam::last=1;
// 从上到下 （实际是从下到上）
for(j=0;j<A[i].size();++j)
{
int u=A[i][j].u;
int c=A[i][j].c;
up[u]=sam::extend(c);
}
// 从下到上 （实际上是从上到下）
sam::last=1;
for(j=A[i].size()-1;j>=0;--j)
{
int u=A[i][j].u;
int c=A[i][j].c;
down[u]=sam::extend(c);
}
}
}
int LCA(int x,int y)
{
while(top[x]!=top[y]) dep[top[x]]>dep[top[y]]?x=fa[top[x]]:y=fa[top[y]];
return dep[x]<dep[y]?x:y;
}
#undef N
};
// 优美的树链剖分
struct node
{
int u,len,ty,tt;
node(int u=0,int len=0,int ty=0,int tt=0):u(u),len(len),ty(ty),tt(tt){}
};
#define N 300003
char S[N];
int ty;
vector<node>A[2];
stack<node>ss;
void insert(int a,int b,int id)
{
int lca=tree::LCA(a,b);
// 向上走
while(tree::dep[a]>tree::dep[lca])
{
int pp=tree::dep[tree::top[a]]<tree::dep[lca]?lca:tree::top[a];
A[id].push_back(node(tree::up[a], tree::dep[a]-max(tree::dep[lca],tree::dep[tree::top[a]])+1,0,a));
a=tree::fa[tree::top[a]];
}
// 向下走
if(a==lca)
{
// 未计算 lca
while(tree::dep[b]>=tree::dep[lca])
{
int pp=tree::dep[tree::top[b]]<tree::dep[lca]?lca:tree::top[b];
ss.push(node(tree::down[pp], tree::dep[b]-max(tree::dep[lca],tree::dep[pp])+1,1,pp));
b=tree::fa[tree::top[b]];
}
}
else
{
while(tree::dep[b]>tree::dep[lca])
{
int pp=tree::dep[tree::top[b]]<tree::dep[lca]?lca:tree::top[b];
ss.push(node(tree::down[pp], tree::dep[b]-max(tree::dep[lca],tree::dep[pp])+1,1,pp));
b=tree::fa[tree::top[b]];
}
}
while(!ss.empty()) A[id].push_back(ss.top()),ss.pop();
}
int main()
{
// setIO("input");
int i,j,n,m;
scanf("%d%s",&n,S+1);
for(i=1;i<=n;++i) tree::val[i]=S[i]-'a';
for(i=1;i<n;++i)
{
int u,v;
scanf("%d%d",&u,&v);
tree::add(u,v);
}
sam::tot=1;
tree::build(); // 建完树剖 + 上/下走定位到后缀自动机的位置
suffix_tree::build(); // 建完后缀树
scanf("%d",&m);
for(i=1;i<=m;++i)
{
int a,b,c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
insert(a,b,0);
insert(c,d,1);
int t1=0,t2=0,lcp=0;
while(t1<A[0].size()&&t2<A[1].size())
{
node a1=A[0][t1], a2=A[1][t2];
int lca;
int uu=a1.u;
int vv=a2.u;
int u=a1.tt;
int v=a2.tt;
int tmp=min(sam::mx[lca=suffix_tree::LCA(uu,vv)], min(a1.len,a2.len));
lcp+=tmp;
if(tmp<min(a1.len,a2.len)) break;
else
{
if(tmp==a1.len) ++t1;
else
{
A[0][t1].u=(A[0][t1].ty==1)?(tree::down[tree::pp[tree::idx[u]+tmp]]):(tree::up[tree::pp[tree::idx[u]-tmp]]);
A[0][t1].tt=(A[0][t1].ty==1)?(tree::pp[tree::idx[u]+tmp]):(tree::pp[tree::idx[u]-tmp]);
A[0][t1].len=A[0][t1].len-tmp;
}
if(tmp==a2.len) ++t2;
else
{
A[1][t2].u=(A[1][t2].ty==1)?(tree::down[tree::pp[tree::idx[v]+tmp]]):(tree::up[tree::pp[tree::idx[v]-tmp]]);
A[1][t2].tt=(A[1][t2].ty==1)?(tree::pp[tree::idx[v]+tmp]):(tree::pp[tree::idx[v]-tmp]);
A[1][t2].len=A[1][t2].len-tmp;
}
}
}
A[0].clear(),A[1].clear();
printf("%d
",lcp);
}
return 0;
}