hiho #1449 : 后缀自动机三·重复旋律6

K = 1..length(S)求出所有长度为K的子串中出现次数最多的子串的出现次数
| endpos(st) |就是st这个状态包含的子串在S中出现的次数
在parent tree上自底向上累加| endpos |,考虑特殊情况：当前状态包含S的前缀，则额外+1，最后统计答案即可

#include <bits/stdc++.h>
#define Cpy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int Maxn = 2000010;
const int maxn = 1000010;
bool vis[Maxn];
int cnt[Maxn], in[Maxn];
struct Suffix_Automata
{
    int maxlen[Maxn], trans[Maxn][26], link[Maxn], Size, Last;
    void init(){Size = Last = 1;}
    inline void Extend(int id)
    {
        int cur = (++Size), p;
        vis[cur] = true;
        maxlen[cur] = maxlen[Last] + 1;
        for (p = Last; p && !trans[p][id]; p = link[p])
            trans[p][id] = cur;
        if (!p)
            link[cur] = 1;
        else
        {
            int q = trans[p][id];
            if (maxlen[q] == maxlen[p] + 1)
                link[cur] = q;
            else
            {
                int clone = (++Size);
                maxlen[clone] = maxlen[p] + 1;
                Cpy(trans[clone], trans[q]);
                link[clone] = link[q];
                for (; p && trans[p][id] == q; p = link[p])
                    trans[p][id] = clone;
                link[cur] = link[q] = clone;
            }
        }
        Last = cur;
    }
} T;
char s[maxn];
void tp()
{
    queue<int> q;
    for (int i = 1; i <= T.Size; i++)
        if (in[i] == 0)
        {
            q.push(i);
            cnt[i] = 1;
        }
    while (!q.empty())
    {
        int p = q.front();
        q.pop();
        int y = T.link[p];
        if (y == 0)
            continue;
        cnt[y] += cnt[p];
        in[y]--;
        if (in[y] == 0)
            q.push(y);
    }
}
int ans[maxn];
int main()
{
    scanf("%s", s);
    T.init();
    int len = strlen(s);
    for (int i = 0; i < len; i++)
        T.Extend(s[i] - 'a');
    for (int i = 1; i <= T.Size; i++)
        in[T.link[i]]++;
    for (int i = 1; i < T.Size; i++)
        if (in[i] != 0 && vis[i] == true)
            cnt[i] = 1;
    tp();
    for (int i = 1; i <= T.Size; i++)
        ans[T.maxlen[i]] = max(ans[T.maxlen[i]], cnt[i]);
    for (int i = len - 1; i >= 1; i--)
        ans[i] = max(ans[i], ans[i + 1]);
    for (int i = 1; i <= len; i++)
        printf("%d\n", ans[i]);
    return 0;
}

原文地址：https://www.cnblogs.com/Zeronera/p/11557819.html