题目大意：给一个字符串S，令F(x)表示S的所有长度为x的子串中，出现次数的最大值。F(1)..F(Length(S))
建出SAM, 然后求出Right, 求Right可以按拓扑序dp..Right就是某个点到结束状态的路径数, parent树上last的那一条链都是结束状态...然后用Right去更新答案..
spoj卡常数..一开始用DFS就炸了, 改用BFS就A了..(贴一下丽洁姐的题解: 我们构造S的SAM，那么对于一个节点s,它的长度范围是 [Min(s),Max(s)],同时他的出现次数是|Right(s)|。那么我们用|Right(s)|去更新F(Max(s))的值。 同时最后从大到小依次用F(i)去更新F(i-1)即可。)
-----------------------------------------------------------------------------
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int cn = 26;
const int maxn = 1000009;
bool vis[maxn];
int Right[maxn], ans[maxn], deg[maxn], n = 0, N;
char s[maxn];
struct Node {
Node *ch[cn], *fa;
int len, id;
} pool[maxn], *pt, *root, *last;
Node* newNode(int v) {
memset(pt->ch, 0, sizeof pt->ch);
pt->fa = 0;
pt->len = v;
pt->id = n++;
return pt++;
}
void SAM_init() {
pt = pool;
root = last = newNode(0);
}
void Extend(int c) {
Node *p = last, *np = newNode(p->len + 1);
for(; p && !p->ch[c]; p = p->fa)
p->ch[c] = np;
if(!p)
np->fa = root;
else {
Node* q = p->ch[c];
if(p->len + 1 == q->len)
np->fa = q;
else {
Node* nq = newNode(p->len + 1);
memcpy(nq->ch, q->ch, sizeof q->ch);
nq->fa = q->fa;
q->fa = np->fa = nq;
for(; p && p->ch[c] == q; p = p->fa)
p->ch[c] = nq;
}
}
last = np;
}
struct edge {
int to;
edge* next;
} E[maxn], *Pt = E, *head[maxn];
void AddEdge(int u, int v) {
deg[Pt->to = v]++; Pt->next = head[u]; head[u] = Pt++;
}
void SAM_build() {
scanf("%s", s);
N = strlen(s);
for(int i = 0; i < N; i++)
Extend(s[i] - ‘a‘);
}
queue<Node*> q;
queue<int> Q;
void ADDEDGE() {
memset(deg, 0, sizeof deg);
memset(vis, 0, sizeof vis);
q.push(root);
vis[root->id] = true;
while(!q.empty()) {
Node* t = q.front(); q.pop();
for(int i = 0; i < cn; i++) if(t->ch[i]) {
AddEdge(t->ch[i]->id, t->id);
if(!vis[t->ch[i]->id]) {
q.push(t->ch[i]);
vis[t->ch[i]->id] = true;
}
}
}
}
void getRight() {
memset(Right, 0, sizeof Right);
for(Node* t = last; t; t = t->fa)
Right[t->id] = 1;
Q.push(last->id);
while(!Q.empty()) {
int x = Q.front(); Q.pop();
for(edge* e = head[x]; e; e = e->next) {
Right[e->to] += Right[x];
if(!--deg[e->to])
Q.push(e->to);
}
}
}
void update() {
memset(vis, 0, sizeof vis);
q.push(root);
vis[root->id] = true;
while(!q.empty()) {
Node* t = q.front(); q.pop();
ans[t->len] = max(ans[t->len], Right[t->id]);
for(int i = 0; i < cn; i++) if(t->ch[i] && !vis[t->ch[i]->id]) {
q.push(t->ch[i]);
vis[t->ch[i]->id] = true;
}
}
}
void solve() {
getRight();
update();
for(int i = N; --i; )
ans[i] = max(ans[i], ans[i + 1]);
for(int i = 1; i <= N; i++)
printf("%d\n", ans[i]);
}
int main() {
SAM_init();
SAM_build();
ADDEDGE();
solve();
return 0;
}
-----------------------------------------------------------------------------
SPOJ8222 Substrings( 后缀自动机 + dp )
原文：http://www.cnblogs.com/JSZX11556/p/4970212.html
