[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
")

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 = '';
    }
    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;
}