后缀自动机
记录一下这几天刷的后缀自动机的题目
Glass Beads UVA - 719
求 \(S\) 循环同构的最小表示
将 \(S\) copy 成 \(SS\) 后建出后缀自动机，按字典序走 \(n\) 步即可
代码
```cpp const int N = 4e4 + 7; char s[N];namespace SAM {
int ch[N][26], len[N], link[N];
int last, tol, n;
void init() {
len[tol = last = 0] = 0; link[0] = -1;
rep (i, 0, 26) ch[0][i] = 0;
tol++;
}
int newnode() {
int p = tol++;
rep (i, 0, 26) ch[p][i] = 0;
len[p] = link[p] = 0;
return p;
}
void extend(int c) {
int cur = newnode();
len[cur] = len[last] + 1;
int p = last;
while (p != -1 && !ch[p][c]) {
ch[p][c] = cur;
p = link[p];
}
if (p == -1) {
link[cur] = 0;
} else {
int q = ch[p][c];
if (len[p] + 1 == len[q]) {
link[cur] = q;
} else {
int clone = newnode();
len[clone] = len[p] + 1;
memcpy(ch[clone], ch[q], sizeof(ch[q]));
link[clone] = link[q];
while (p != -1 && ch[p][c] == q) {
ch[p][c] = clone;
p = link[p];
}
link[q] = link[cur] = clone;
}
}
last = cur;
}
int dfs(int u, int l) {
if (l == n) return len[u];
rep (i, 0, 26) if (ch[u][i]) return dfs(ch[u][i], l + 1);
}
void solve() {
scanf("%s", s);
n = strlen(s);
init();
rep (l, 0, 2) rep (i, 0, n) extend(s[i] - 'a');
printf("%d\n", dfs(0, 0) - n + 1);
}
}
int main() {
ifdef LOCAL
freopen("ans.out", "w", stdout);
endif
int T;
scanf("%d", &T);
while (T--) SAM::solve();
ifdef LOCAL
printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
endif
return 0;
}
</details>
[Longest Common Substring SPOJ - LCS](https://vjudge.net/problem/SPOJ-LCS)
求两个串的最长公共子串
对第一个串建后缀自动机，第二个串在自动机上匹配，若匹配上了则长度加一，并且转移到下一个状态，匹配失败则走fail节点，更新 $len$，每次都更新一下答案即可
<details>
<summary>代码</summary>
```cpp
const int N = 5e5 + 7;
namespace SAM {
int ch[N][26], len[N], link[N], sz[N];
int last, tol;
void init() {
len[tol = last = 0] = 0; link[0] = -1;
rep (i, 0, 26) ch[0][i] = 0;
tol++;
}
int newnode() {
int p = tol++;
rep (i, 0, 26) ch[p][0] = 0;
len[p] = link[p] = sz[p] = 0;
return p;
}
void extend(int c) {
int cur = newnode();
len[cur] = len[last] + 1;
sz[cur] = 1;
int p = last;
while (p != -1 && !ch[p][c]) {
ch[p][c] = cur;
p = link[p];
}
if (p == -1) {
link[cur] = 0;
} else {
int q = ch[p][c];
if (len[p] + 1 == len[q]) {
link[cur] = q;
} else {
int clone = newnode();
len[clone] = len[p] + 1;
memcpy(ch[clone], ch[q], sizeof(ch[q]));
link[clone] = link[q];
while (p != -1 && ch[p][c] == q) {
ch[p][c] = clone;
p = link[p];
}
link[q] = link[cur] = clone;
}
}
last = cur;
}
void solve(char *s) {
int n = strlen(s);
int cur = 0;
int ans = 0, l = 0;
rep (i, 0, n) {
int d = s[i] - 'a';
if (ch[cur][d]) {
cur = ch[cur][d];
l++;
} else {
while (cur != -1 && !ch[cur][d]) cur = link[cur];
if (cur == -1) cur = l = 0;
else l = len[cur] + 1, cur = ch[cur][d];
}
chkmax(ans, l);
}
printf("%d\n", ans);
}
}
char s[N], t[N];
int main() {
#ifdef LOCAL
freopen("ans.out", "w", stdout);
#endif
scanf("%s%s", s, t);
int n = strlen(s);
SAM::init();
rep (i, 0, n) SAM::extend(s[i] - 'a');
SAM::solve(t);
#ifdef LOCAL
printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
#endif
return 0;
}
Substrings SPOJ - NSUBSTR
求每个长度right集合最大的大小
right集合大小通过fail树进行dfs或拓扑序求出，然后用right集合大小更新状态里最长串
最后从后往前依次用 \(f_{i+1}\) 更新 \(f_i\) 即可
代码
```cpp const int N = 5e5 + 7, E = 5e5 + 7; int f[N]; namespace SAM { int ch[N][26], len[N], link[N], cnt[N]; int last, tol; void init() { len[tol = last = 0] = 0; link[0] = -1; rep (i, 0, 26) ch[0][i] = 0; tol++; } int newnode() { int p = tol++; rep (i, 0, 26) ch[p][0] = 0; len[p] = link[p] = cnt[p] = 0; return p; } void extend(int c) { int cur = newnode(); len[cur] = len[last] + 1; cnt[cur] = 1; int p = last; while (p != -1 && !ch[p][c]) { ch[p][c] = cur; p = link[p]; } if (p == -1) { link[cur] = 0; } else { int q = ch[p][c]; if (len[p] + 1 == len[q]) { link[cur] = q; } else { int clone = newnode(); len[clone] = len[p] + 1; memcpy(ch[clone], ch[q], sizeof(ch[q])); link[clone] = link[q]; while (p != -1 && ch[p][c] == q) { ch[p][c] = clone; p = link[p]; } link[q] = link[cur] = clone; } } last = cur; } Edg void dfs(int u) { es (u, i, v) dfs(v), cnt[u] += cnt[v]; chkmax(f[len[u]], cnt[u]); } void count() { rep (i, 1, tol) addd(link[i], i); dfs(0); } }char s[N];
int n;
int main() {
ifdef LOCAL
freopen("ans.out", "w", stdout);
endif
SAM::init();
scanf("%s", s);
n = strlen(s);
rep (i, 0, n) SAM::extend(s[i] - 'a');
SAM::count();
per (i, 1, n) chkmax(f[i], f[i + 1]);
rep (i, 1, n + 1) printf("%d\n", f[i]);
ifdef LOCAL
printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
endif
return 0;
}
</details>
[Longest Common Substring II SPOJ - LCS2](https://vjudge.net/problem/SPOJ-LCS2)
求多个串之间的最长公共子串
首先对第一个串建后缀自动机，然后对后面每一个串进行匹配，保存在每个状态上能匹配长度的最大值，再用拓扑序更新fail能匹配的最大值（因为当前节点能匹配的长度是这个状态的一个后缀，fail系欸但也是该状态的后缀，那么当前状态能匹配上的长度fail也能匹配上）
然后分别取min即可（因为要满足每个串都能匹配上，那么是最短的限制了它们能匹配的最长长度）
<details>
<summary>代码</summary>
```cpp
const int N = 5e5 + 7;
int ch[N][26], len[N], link[N], cnt[N];
int last, tol;
void init() {
len[tol = last = 0] = 0; link[0] = -1;
rep (i, 0, 26) ch[0][i] = 0;
tol++;
}
int newnode() {
int p = tol++;
rep (i, 0, 26) ch[p][0] = 0;
len[p] = link[p] = cnt[p] = 0;
return p;
}
void extend(int c) {
int cur = newnode();
len[cur] = len[last] + 1;
cnt[cur] = 1;
int p = last;
while (p != -1 && !ch[p][c]) {
ch[p][c] = cur;
p = link[p];
}
if (p == -1) {
link[cur] = 0;
} else {
int q = ch[p][c];
if (len[p] + 1 == len[q]) {
link[cur] = q;
} else {
int clone = newnode();
len[clone] = len[p] + 1;
memcpy(ch[clone], ch[q], sizeof(ch[q]));
link[clone] = link[q];
while (p != -1 && ch[p][c] == q) {
ch[p][c] = clone;
p = link[p];
}
link[q] = link[cur] = clone;
}
}
last = cur;
}
int c[N], a[N];
int mx[N], ans[N];
char s[N];
int main() {
#ifdef LOCAL
freopen("ans.out", "w", stdout);
#endif
memset(ans, 0x3f, sizeof(ans));
init();
scanf("%s", s);
for (int i = 0; s[i]; i++) extend(s[i] - 'a');
rep (i, 0, tol) c[len[i]]++;
rep (i, 1, tol) c[i] += c[i - 1];
rep (i, 0, tol) a[--c[len[i]]] = i;
while (~scanf("%s", s)) {
int l = 0, p = 0;
for (int i = 0; s[i]; i++) {
int d = s[i] - 'a';
while (~p && !ch[p][d]) p = link[p], l = len[p];
if (p == -1) p = l = 0;
else ++l, p = ch[p][d], chkmax(mx[p], l);
}
per (i, 0, tol) {
int p = a[i];
if (~link[p]) chkmax(mx[link[p]], std::min(len[link[p]], mx[p]));
chkmin(ans[p], mx[p]);
mx[p] = 0;
}
}
int res = 0;
rep (i, 0, tol) chkmax(res, ans[i]);
printf("%d\n", res);
#ifdef LOCAL
printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
#endif
return 0;
}
Lexicographical Substring Search SPOJ - SUBLEX
求出字典序第 \(k\) 小的子串
每个状态的right集合为 \(1\)，再求出每个状态往 \(trans\) 边走的路径数，通过路径数类似于主席树求第 \(k\) 大的方法dfs即可
求路径数为 \(f_u = right_u + \sum \limits_{u \to v} f_v\)
代码
```cpp const int N = 2e5 + 7; int ch[N][26], len[N], link[N], cnt[N], f[N]; int last, tol; void init() { len[tol = last = 0] = 0; link[0] = -1; rep (i, 0, 26) ch[0][i] = 0; tol++; } int newnode() { int p = tol++; rep (i, 0, 26) ch[p][0] = 0; len[p] = link[p] = cnt[p] = 0; return p; } void extend(int c) { int cur = newnode(); len[cur] = len[last] + 1; cnt[cur] = 1; int p = last; while (p != -1 && !ch[p][c]) { ch[p][c] = cur; p = link[p]; } if (p == -1) { link[cur] = 0; } else { int q = ch[p][c]; if (len[p] + 1 == len[q]) { link[cur] = q; } else { int clone = newnode(); len[clone] = len[p] + 1; memcpy(ch[clone], ch[q], sizeof(ch[q])); link[clone] = link[q]; while (p != -1 && ch[p][c] == q) { ch[p][c] = clone; p = link[p]; } link[q] = link[cur] = clone; } } last = cur; }void dfs(int u, int x) {
if (!x) return;
rep (i, 0, 26) {
int ne = ch[u][i];
if (ne <= 0) continue;
if (f[ne] >= x) {
putchar('a' + i);
dfs(ne, x - 1);
return;
}
x -= f[ne];
}
}
char s[N];
int a[N], c[N];
int main() {
ifdef LOCAL
freopen("ans.out", "w", stdout);
endif
init();
scanf("%s", s);
for (int i = 0; s[i]; i++) extend(s[i] - 'a');
rep (i, 0, tol) c[len[i]]++;
rep (i, 1, tol) c[i] += c[i - 1];
rep (i, 0, tol) a[--c[len[i]]] = i;
per (i, 0, tol) {
int p = a[i];
f[p] = 1;
rep (j, 0, 26) if (ch[p][j]) f[p] += f[ch[p][j]];
}
int T;
scanf("%d", &T);
while (T--) {
int x;
scanf("%d", &x);
dfs(0, x);
puts("");
}
ifdef LOCAL
printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
endif
return 0;
}
</details>
[HDU4622 Reincarnation](http://acm.hdu.edu.cn/showproblem.php?pid=4622)
求区间本质不同子串个数
离线询问，按左端点排序，左端点相同的建后缀自动机，复杂度为 $O(n^2)$
<details>
<summary>代码</summary>
```cpp
const int N = 2e4 + 7;
const int Q = 2e4 + 7;
struct Qu {
int l, r, id;
bool operator < (const Qu &p) const {
return l == p.l ? r < p.r : l < p.l;
}
} query[N];
int ch[N][26], len[N], link[N], sz[N];
int last, tol;
int ans;
void init() {
len[0] = 0; link[0] = -1;
tol = 0, last = 0, ans = 0;
rep (i, 0, 26) ch[0][i] = 0;
tol++;
}
int newnode() {
int p = tol++;
rep (i, 0, 26) ch[p][i] = 0;
len[p] = link[p] = sz[p] = 0;
return p;
}
void extend(int c) {
int cur = newnode();
len[cur] = len[last] + 1;
sz[cur] = 1;
int p = last;
while (p != -1 && !ch[p][c]) {
ch[p][c] = cur;
p = link[p];
}
if (p == -1) {
link[cur] = 0;
} else {
int q = ch[p][c];
if (len[p] + 1 == len[q]) {
link[cur] = q;
} else {
int clone = newnode();
len[clone] = len[p] + 1;
memcpy(ch[clone], ch[q], sizeof(ch[q]));
link[clone] = link[q];
while (p != -1 && ch[p][c] == q) {
ch[p][c] = clone;
p = link[p];
}
link[q] = link[cur] = clone;
}
}
last = cur;
ans += len[cur] - len[link[cur]];
}
char s[N];
int res[Q];
void solve() {
scanf("%s", s + 1);
int n = strlen(s + 1);
int q;
scanf("%d", &q);
rep (i, 1, q + 1) scanf("%d%d", &query[i].l, &query[i].r), query[i].id = i;
std::sort(query + 1, query + q + 1);
query[0].l = -1;
rep (i, 1, q + 1) {
if (query[i].l != query[i - 1].l) {
init();
rep (j, query[i].l, query[i].r + 1) extend(s[j] - 'a');
res[query[i].id] = ans;
} else {
rep (j, query[i - 1].r + 1, query[i].r + 1) extend(s[j] - 'a');
res[query[i].id] = ans;
}
}
rep (i, 1, q + 1) printf("%d\n", res[i]);
}
int main() {
#ifdef LOCAL
freopen("ans.out", "w", stdout);
#endif
int T;
scanf("%d", &T);
while (T--) solve();
#ifdef LOCAL
printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
#endif
return 0;
}
HDU4641 K-string
插入字符或求right集合不小于 \(K\) 的子串个数
离线，建出后缀自动机后，一个right集合对答案有贡献对于时间来说是一个后缀，具体为从right集合第 \(k\) 大开始贡献
在fail树上从下至上进行线段树合并，查询线段树第 \(k\) 大即可
代码
```cpp const int N = 5e5 + 7; namespace SAM { int ch[N][26], len[N], fail[N], id[N]; int last, tol; void init() { len[0] = 0; fail[0] = -1; tol = 0, last = 0; rep (i, 0, 26) ch[0][i] = 0; tol++; } int newnode() { int p = tol++; rep (i, 0, 26) ch[p][i] = 0; len[p] = fail[p] = id[p] = 0; return p; } void extend(int c, int n) { int cur = newnode(); len[cur] = len[last] + 1; assert(len[cur] == n); id[n] = cur; int p = last; for (; p != -1 && !ch[p][c]; p = fail[p]) ch[p][c] = cur; if (p == -1) { fail[cur] = 0; } else { int q = ch[p][c]; if (len[p] + 1 == len[q]) { fail[cur] = q; } else { int clone = newnode(); len[clone] = len[p] + 1; memcpy(ch[clone], ch[q], sizeof(ch[q])); fail[clone] = fail[q]; for ( ; p != -1 && ch[p][c] == q; p = fail[p]) ch[p][c] = clone; fail[q] = fail[cur] = clone; } } last = cur; } } struct Tree { int l, r, sum; void clear() { l = r = sum = 0; } } tree[N * 16]; int tol; void init() { rep (i, 0, tol + 1) tree[i].clear(); tol = 0; } void update(int &p, int l, int r, int pos) { p = ++tol; tree[p].sum++; if (l == r) return; if (pos <= mid) update(lp, l, mid, pos); else update(rp, mid + 1, r, pos); } int merge(int x, int y, int l, int r) { if (!x || !y) return x | y; tree[x].sum = tree[x].sum + tree[y].sum; if (l < r) { tree[x].l = merge(tree[x].l, tree[y].l, l, mid); tree[x].r = merge(tree[x].r, tree[y].r, mid + 1, r); } else { tree[x].l = tree[x].r = 0; } return x; } int query(int p, int l, int r, int k) { if (l == r) return l; if (k <= tree[lp].sum) return query(lp, l, mid, k); return query(rp, mid + 1, r, k - tree[lp].sum); }int n, m, k, que[N * 2], q, a[N], c[N], root[N];
ll ans[N];
char s[N];
void solve() {
scanf("%s", s + 1);
SAM::init();
rep (i, 1, n + 1) SAM::extend(s[i] - 'a', i);
q = 0;
rep (i, 0, m) {
int opt;
scanf("%d", &opt);
if (opt == 2) que[++q] = n;
else {
static char str[3];
scanf("%s", str);
n++;
SAM::extend(str[0] - 'a', n);
}
}
init();
rep (i, 0, SAM::tol) root[i] = 0;
rep (i, 1, n + 1) update(root[SAM::id[i]], 1, n, i), ans[i] = 0;
rep (i, 0, SAM::tol) c[i] = 0;
rep (i, 0, SAM::tol) c[SAM::len[i]]++;
rep (i, 1, SAM::tol) c[i] += c[i - 1];
rep (i, 0, SAM::tol) a[--c[SAM::len[i]]] = i;
per (i, 0, SAM::tol) {
int p = a[i];
if (tree[root[p]].sum >= k) ans[query(root[p], 1, n, k)] += SAM::len[p] - (SAM::fail[p] >= 0 ? SAM::len[SAM::fail[p]] : 0);
if (SAM::fail[p] >= 0) root[SAM::fail[p]] = merge(root[SAM::fail[p]], root[p], 1, n);
}
rep (i, 1, n + 1) ans[i] += ans[i - 1];
rep (i, 1, q + 1) printf("%lld\n", ans[que[i]]);
}
int main() {
ifdef LOCAL
freopen("ans.out", "w", stdout);
endif
while (~scanf("%d%d%d", &n, &m, &k)) solve();
ifdef LOCAL
printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
endif
return 0;
}
</details>
[BZOJ 3998: [TJOI2015]弦论](http://www.lydsy.com/JudgeOnline/problem.php?id=3998)
T=0时right集合为 $1$，T=1时right集合为实际大小，拓扑序求出
再dfs得到第 $k$ 小字典序
<details>
<summary>代码</summary>
```cpp
const int N = 1e6 + 7;
int ch[N][26], len[N], fail[N], f[N], right[N];
char s[N];
int last, tol;
void init() {
len[tol = last = 0] = 0; fail[0] = -1;
rep (i, 0, 26) ch[0][i] = 0;
tol++;
}
int newnode() {
int p = tol++;
rep (i, 0, 26) ch[p][i] = 0;
len[p] = fail[p] = right[p] = f[p] = 0;
return p;
}
void extend(int c) {
int cur = newnode();
len[cur] = len[last] + 1;
right[cur] = 1;
int p = last;
while (p != -1 && !ch[p][c]) {
ch[p][c] = cur;
p = fail[p];
}
if (p == -1) {
fail[cur] = 0;
} else {
int q = ch[p][c];
if (len[p] + 1 == len[q]) {
fail[cur] = q;
} else {
int clone = newnode();
len[clone] = len[p] + 1;
memcpy(ch[clone], ch[q], sizeof(ch[q]));
fail[clone] = fail[q];
while (p != -1 && ch[p][c] == q) {
ch[p][c] = clone;
p = fail[p];
}
fail[q] = fail[cur] = clone;
}
}
last = cur;
}
void dfs(int cur, int k) {
if (k <= right[cur]) return;
k -= right[cur];
rep (i, 0, 26) if (ch[cur][i] > 0) {
int ne = ch[cur][i];
if (k <= f[ne]) {
putchar('a' + i);
dfs(ne, k);
return;
}
k -= f[ne];
}
}
int a[N], c[N];
int main() {
#ifdef LOCAL
freopen("ans.out", "w", stdout);
#endif
scanf("%s", s);
init();
for (int i = 0; s[i]; i++) extend(s[i] - 'a');
int opt, k;
scanf("%d%d", &opt, &k);
rep (i, 0, tol) c[len[i]]++;
rep (i, 1, tol) c[i] += c[i - 1];
rep (i, 0, tol) a[--c[len[i]]] = i;
per (i, 0, tol) {
int p = a[i];
if (opt) {
if (fail[p] >= 0) right[fail[p]] += right[p];
} else {
right[i] = 1;
}
}
right[0] = 0;
per (i, 0, tol) {
int p = a[i];
f[p] = right[p];
rep (j, 0, 26) if (ch[p][j] >= 0) f[p] += f[ch[p][j]];
}
if (f[0] < k) return puts("-1"), 0;
dfs(0, k);
puts("");
#ifdef LOCAL
printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
#endif
return 0;
}
BZOJ 4516: [Sdoi2016]生成魔咒
动态插入字符，求每个时刻的本质不同子串的个数
每次插入字符后答案加上 \(len_{cur}-len_{fail_{cur}}\) 即可
代码
```cpp const int N = 2e5 + 7; namespace SAM { std::mapchar s[N];
int main() {
ifdef LOCAL
freopen("ans.out", "w", stdout);
endif
SAM::init();
int n = _();
rep (i, 0, n) {
int x = _();
SAM::extend(x), printf("%lld\n", SAM::ans);
}
ifdef LOCAL
printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
endif
return 0;
}
</details>
