Omg, I don't know for sure but second code looks like counting sort from suffix array. Can you refer to some comments or submissions with it, please?

Ok, I think, I understood what this code states for. This is very weird way to do this thing but anyway.

==1==

for (int i = 0,p = S; i < n; ++i)
	++r[p = ch[p][s[i] - 'a']];

You want to count for every state of suffix automata how many times does it appear in string. It will be array r. You initialize it with 1 in every state which corresponds to some prefix of string.

==2==

static int b[N],t[N];
for (int i = 1; i <= cnt; ++i) ++b[l[i]];
for (int i = 1; i <= n; ++i) b[i] += b[i - 1];
for (int i = 1; i <= cnt; ++i) t[b[l[i]]--] = i;

Actually in this part you obtain lexicographical order of states. It is equivalent to counting sort by lengths of states. At the end array t contains topological sort of states.

==3==

for (int i = cnt; i; --i) r[fa[t[i]]] += r[t[i]];

Finally you finish calculating your array r by pushing initial values through the suffix links of automaton in reversed topological sort order. This procedure is described on e-maxx, I suppose.

==4==

Here is equivalent simplificated code for such thing.

int occur[max_size]; // occur[state] = 1 if state is prefix of string
vector<int> state_with_len[max_len];
for(int state = 0; state < state_count; state++)
	state_with_len[len[state]].push_back(state);
// If you traverse states from bigger length to smallest you will have
// Reversed topological order. The same which you use in dfs actually. 
for(int length = max_len - 1; length > 0; length--) 
	for(auto state: state_with_len[length])
		occur[link[state]] += occur[state];

P.S. I suppose this is Chinese code? Such crazy one :)

I wrote about endpos concept here. Please write if more clarification is needed.