SPOJ：NSUBSTR - Substrings

题面

字符串$ S \(最多包含\) 25 \(万个小写拉丁字母。我们将\) F(x) \(定义为长度为\) x \(的某些字符串出现在\) s \(中的最大次数。例如，对于字符串\) “ababa”\(，\)F(3) \(将为\) 2\(，因为存在两次出现的字符串\) “aba”\(。您的任务是为每个\) i $输出 \(F(i)\)，以便$ 1 <= i < = |S|$

Sol

\(sam\)

直接求一下每个\(endpos(right)\)集合的子串出现次数

然后就没了

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

template <class Int>
IL void Input(RG Int &x){
    RG int z = 1; RG char c = getchar(); x = 0;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    x *= z;
}

const int maxn(5e5 + 5);

int n, trans[26][maxn], fa[maxn], len[maxn], tot = 1, last = 1;
int id[maxn], t[maxn], size[maxn], ans[maxn];
char s[maxn];

IL void Extend(RG int c){
    RG int p = last, np = ++tot; last = tot;
    len[np] = len[p] + 1, size[np] = 1;
    while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
    if(!p) fa[np] = 1;
    else{
        RG int q = trans[c][p];
        if(len[q] == len[p] + 1) fa[np] = q;
        else{
            RG int nq = ++tot;
            fa[nq] = fa[q], len[nq] = len[p] + 1;
            for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
            fa[q] = fa[np] = nq;
            while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
        }
    }
}

int main(RG int argc, RG char* argv[]){
    scanf(" %s", s), n = strlen(s);
    for(RG int i = 0; i < n; ++i) Extend(s[i] - 'a');
    for(RG int i = 1; i <= tot; ++i) ++t[len[i]];
    for(RG int i = 1; i <= tot; ++i) t[i] += t[i - 1];
    for(RG int i = 1; i <= tot; ++i) id[t[len[i]]--] = i;
    for(RG int i = tot; i; --i){
        size[fa[id[i]]] += size[id[i]];
        ans[len[id[i]]] = max(ans[len[id[i]]], size[id[i]]);
    }
    for(RG int i = tot; i; --i) ans[i] = max(ans[i], ans[i + 1]);
    for(RG int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
    return 0;
}