[模板]后缀自动机
题面:
给一个字符串 \(S\) ,求出 \(S\) 的所有出现次数不为 \(1\) 的子串的出现次数乘上该子串长度的最大值
法一:SAM(由于本蒟蒻太菜现在还不会)
咕咕咕...
法二:SA
先后缀排序, 求出 \(Height\) 数组,然后弄好ST表,从小到大处理每一个前缀,统计完一个前缀的答案后,就删掉它,然后再继续操作左右两个裂开的区间,代码写的很清楚,这里不在赘述。
SA的做法还是挺巧妙的。
#include <set>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <assert.h>
#include <algorithm>
using namespace std;
#define fir first
#define sec second
#define pb push_back
#define mp make_pair
#define LL long long
#define INF (0x3f3f3f3f)
#define mem(a, b) memset(a, b, sizeof (a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define Debug(x) cout << #x << " = " << x << endl
#define tralve(i, x) for (register int i = head[x]; i; i = nxt[i])
#define For(i, a, b) for (register int (i) = (a); (i) <= (b); ++ (i))
#define Forr(i, a, b) for (register int (i) = (a); (i) >= (b); -- (i))
#define file(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
#define ____ debug("go\n")
namespace io {
static char buf[1<<21], *pos = buf, *end = buf;
inline char getc()
{ return pos == end && (end = (pos = buf) + fread(buf, 1, 1<<21, stdin), pos == end) ? EOF : *pos ++; }
inline int rint() {
register int x = 0, f = 1;register char c;
while (!isdigit(c = getc())) if (c == '-') f = -1;
while (x = (x << 1) + (x << 3) + (c ^ 48), isdigit(c = getc()));
return x * f;
}
inline LL rLL() {
register LL x = 0, f = 1; register char c;
while (!isdigit(c = getc())) if (c == '-') f = -1;
while (x = (x << 1ll) + (x << 3ll) + (c ^ 48), isdigit(c = getc()));
return x * f;
}
inline void rstr(char *str) {
while (isspace(*str = getc()));
while (!isspace(*++str = getc()))
if (*str == EOF) break;
*str = '\0';
}
template<typename T>
inline bool chkmin(T &x, T y) { return x > y ? (x = y, 1) : 0; }
template<typename T>
inline bool chkmax(T &x, T y) { return x < y ? (x = y, 1) : 0; }
}
using namespace io;
const int N = 1e6 + 2;
namespace SA {
char str[N];
int n, sa[N], rk[N], ht[N], M;
void Rsort(int *buc, int *tp) {
fill(buc, buc + 1 + M, 0);
For (i, 1, n) buc[rk[i]] ++;
For (i, 1, M) buc[i] += buc[i - 1];
Forr (i, n, 1) sa[ buc[ rk[ tp[i] ] ] -- ] = tp[i];
}
void SuffixSort() {
static int buc[N], tp[N];
M = 76;
For (i, 1, n) rk[i] = str[i - 1] - '0', tp[i] = i;
Rsort(buc, tp);
for (register int len = 1, cnt = 0; cnt < n; len <<= 1, M = cnt) {
cnt = 0;
For (i, n - len + 1, n) tp[++cnt] = i;
For (i, 1, n) if (sa[i] - len > 0) tp[++cnt] = sa[i] - len;
Rsort(buc, tp); swap(rk, tp); rk[sa[1]] = cnt = 1;
For (i, 2, n) rk[sa[i]] = (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + len] == tp[sa[i - 1] + len] ? cnt : ++cnt);
}
}
void getHt() {
int len = 0;
For (i, 1, n) {
if (len) len --;
int j = sa[rk[i] - 1];
while (str[j + len - 1] == str[i + len - 1]) len ++;
ht[rk[i]] = len;
}
}
#define pii pair<int, int>
int Log[N];
pii st[N][22];
void STinit() {
Log[0] = -1;
For (i, 1, n) Log[i] = Log[i >> 1] + 1, st[i][0] = mp(ht[i], i);
for (register int i = 1; n >> i; ++ i)
for (register int j = 1; j + (1 << i) - 1 <= n; ++ j)
st[j][i] = min(st[j][i - 1], st[j + (1 << i - 1)][i - 1]);
}
pii query(int l, int r) {
int len = Log[r - l + 1];
return min(st[l][len], st[r - (1 << len) + 1][len]);
}
#undef pii
}
int ans = 0;
void solve(int l, int r) {
if (l > r) return;
pair<int, int> tmp = SA::query(l, r);
chkmax(ans, (r - l + 2) * tmp.fir);
solve(l, tmp.sec - 1);
solve(tmp.sec + 1, r);
}
int main() {
#ifndef ONLINE_JUDGE
file("SA");
#endif
scanf("%s", SA::str);
SA::n = strlen(SA::str);
SA::SuffixSort();
SA::getHt();
SA::STinit();
solve(1, SA::n);
cout << ans << endl;
}