This article is a stub and doesn’t contain any descriptions. For a description of the algorithm, refer to other sources, such as Algorithms on Strings, Trees, and Sequences by Dan Gusfield.
This algorithm builds a suffix tree for a given string $s$ of length $n$ in $O(n\log(k))$) time, where $k$ is the size of the alphabet (if $k$ is considered to be a constant, the asymptotic behavior is linear).
The input to the algorithm are the string $s$ and its length $n$, which are passed as global variables.
The main function build_tree builds a suffix tree. It is stored as an array of structures node, where node[0] is the root of the tree.
In order to simplify the code, the edges are stored in the same structures: for each vertex its structure node stores the information about the edge between it and its parent. Overall each node stores the following information:
(l, r)- left and right boundaries of the substrings[l..r-1]which correspond to the edge to this node,par- the parent node,link- the suffix link,next- the list of edges going out from this node.
string s;
int n;
struct node {
int l, r, par, link;
map<char,int> next;
node (int l=0, int r=0, int par=-1)
: l(l), r(r), par(par), link(-1) {}
int len() { return r - l; }
int &get (char c) {
if (!next.count(c)) next[c] = -1;
return next[c];
}
};
node t[MAXN];
int sz;
struct state {
int v, pos;
state (int v, int pos) : v(v), pos(pos) {}
};
state ptr (0, 0);
state go (state st, int l, int r) {
while (l < r)
if (st.pos == t[st.v].len()) {
st = state (t[st.v].get( s[l] ), 0);
if (st.v == -1) return st;
}
else {
if (s[ t[st.v].l + st.pos ] != s[l])
return state (-1, -1);
if (r-l < t[st.v].len() - st.pos)
return state (st.v, st.pos + r-l);
l += t[st.v].len() - st.pos;
st.pos = t[st.v].len();
}
return st;
}
int split (state st) {
if (st.pos == t[st.v].len())
return st.v;
if (st.pos == 0)
return t[st.v].par;
node v = t[st.v];
int id = sz++;
t[id] = node (v.l, v.l+st.pos, v.par);
t[v.par].get( s[v.l] ) = id;
t[id].get( s[v.l+st.pos] ) = st.v;
t[st.v].par = id;
t[st.v].l += st.pos;
return id;
}
int get_link (int v) {
if (t[v].link != -1) return t[v].link;
if (t[v].par == -1) return 0;
int to = get_link (t[v].par);
return t[v].link = split (go (state(to,t[to].len()), t[v].l + (t[v].par==0), t[v].r));
}
void tree_extend (int pos) {
for(;;) {
state nptr = go (ptr, pos, pos+1);
if (nptr.v != -1) {
ptr = nptr;
return;
}
int mid = split (ptr);
int leaf = sz++;
t[leaf] = node (pos, n, mid);
t[mid].get( s[pos] ) = leaf;
ptr.v = get_link (mid);
ptr.pos = t[ptr.v].len();
if (!mid) break;
}
}
void build_tree() {
sz = 1;
for (int i=0; i<n; ++i)
tree_extend (i);
}
Compressed Implementation
This compressed implementation was proposed by freopen.
const int N=1000000,INF=1000000000;
string a;
int t[N][26],l[N],r[N],p[N],s[N],tv,tp,ts,la;
void ukkadd (int c) {
suff:;
if (r[tv]<tp) {
if (t[tv][c]==-1) { t[tv][c]=ts; l[ts]=la;
p[ts++]=tv; tv=s[tv]; tp=r[tv]+1; goto suff; }
tv=t[tv][c]; tp=l[tv];
}
if (tp==-1 || c==a[tp]-'a') tp++; else {
l[ts+1]=la; p[ts+1]=ts;
l[ts]=l[tv]; r[ts]=tp-1; p[ts]=p[tv]; t[ts][c]=ts+1; t[ts][a[tp]-'a']=tv;
l[tv]=tp; p[tv]=ts; t[p[ts]][a[l[ts]]-'a']=ts; ts+=2;
tv=s[p[ts-2]]; tp=l[ts-2];
while (tp<=r[ts-2]) { tv=t[tv][a[tp]-'a']; tp+=r[tv]-l[tv]+1;}
if (tp==r[ts-2]+1) s[ts-2]=tv; else s[ts-2]=ts;
tp=r[tv]-(tp-r[ts-2])+2; goto suff;
}
}
void build() {
ts=2;
tv=0;
tp=0;
fill(r,r+N,(int)a.size()-1);
s[0]=1;
l[0]=-1;
r[0]=-1;
l[1]=-1;
r[1]=-1;
memset (t, -1, sizeof t);
fill(t[1],t[1]+26,0);
for (la=0; la<(int)a.size(); ++la)
ukkadd (a[la]-'a');
}
Same code with comments:
const int N=1000000, // maximum possible number of nodes in suffix tree
INF=1000000000; // infinity constant
string a; // input string for which the suffix tree is being built
int t[N][26], // array of transitions (state, letter)
l[N], // left...
r[N], // ...and right boundaries of the substring of a which correspond to incoming edge
p[N], // parent of the node
s[N], // suffix link
tv, // the node of the current suffix (if we're mid-edge, the lower node of the edge)
tp, // position in the string which corresponds to the position on the edge (between l[tv] and r[tv], inclusive)
ts, // the number of nodes
la; // the current character in the string
void ukkadd(int c) { // add character s to the tree
suff:; // we'll return here after each transition to the suffix (and will add character again)
if (r[tv]<tp) { // check whether we're still within the boundaries of the current edge
// if we're not, find the next edge. If it doesn't exist, create a leaf and add it to the tree
if (t[tv][c]==-1) {t[tv][c]=ts;l[ts]=la;p[ts++]=tv;tv=s[tv];tp=r[tv]+1;goto suff;}
tv=t[tv][c];tp=l[tv];
} // otherwise just proceed to the next edge
if (tp==-1 || c==a[tp]-'a')
tp++; // if the letter on the edge equal c, go down that edge
else {
// otherwise split the edge in two with middle in node ts
l[ts]=l[tv];r[ts]=tp-1;p[ts]=p[tv];t[ts][a[tp]-'a']=tv;
// add leaf ts+1. It corresponds to transition through c.
t[ts][c]=ts+1;l[ts+1]=la;p[ts+1]=ts;
// update info for the current node - remember to mark ts as parent of tv
l[tv]=tp;p[tv]=ts;t[p[ts]][a[l[ts]]-'a']=ts;ts+=2;
// prepare for descent
// tp will mark where are we in the current suffix
tv=s[p[ts-2]];tp=l[ts-2];
// while the current suffix is not over, descend
while (tp<=r[ts-2]) {tv=t[tv][a[tp]-'a'];tp+=r[tv]-l[tv]+1;}
// if we're in a node, add a suffix link to it, otherwise add the link to ts
// (we'll create ts on next iteration).
if (tp==r[ts-2]+1) s[ts-2]=tv; else s[ts-2]=ts;
// add tp to the new edge and return to add letter to suffix
tp=r[tv]-(tp-r[ts-2])+2;goto suff;
}
}
void build() {
ts=2;
tv=0;
tp=0;
fill(r,r+N,(int)a.size()-1);
// initialize data for the root of the tree
s[0]=1;
l[0]=-1;
r[0]=-1;
l[1]=-1;
r[1]=-1;
memset (t, -1, sizeof t);
fill(t[1],t[1]+26,0);
// add the text to the tree, letter by letter
for (la=0; la<(int)a.size(); ++la)
ukkadd (a[la]-'a');
}