SPOJ8222 NSUBSTR - Substrings(后缀自动机)

You are given a string S which consists of 250000 lowercase latin letters at most. We define F(x) as the maximal number of times that some string with length x appears in S. For example for string 'ababa' F(3) will be 2 because there is a string 'aba' that occurs twice. Your task is to output F(i) for every i so that 1<=i<=|S|.

Input

String S consists of at most 250000 lowercase latin letters.

Output

Output |S| lines. On the i-th line output F(i).

Example

Input:
ababa

Output:
3
2
2
1
1

Submit solution!

为什么我的后缀自动机写的这么奇怪qwq。。

为什么泥萌都在写桶排？？！！

这题不是dfs一下求出$right$集合来就完了么qwq、、

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int MAXN =  << , INF = 1e9 + ;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
char s[MAXN];
int N;
int f[MAXN], siz[MAXN], fa[MAXN], len[MAXN], ch[MAXN][], root = , last = , tot = ;
void insert(int x) {
    int now = ++tot, pre = last; last = now;
    siz[now] = ;
    len[now] = len[pre] + ;
    for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
    if(!pre) fa[now] = root;
    else {
        int q = ch[pre][x];
        if(len[q] == len[pre] + ) fa[now] = q;
        else {
            int nows = ++tot;
            memcpy(ch[nows], ch[q], sizeof(ch[q]));
            fa[nows] = fa[q]; fa[q] = fa[now] = nows;
            len[nows] = len[pre] + ;
            for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nows;
        }
    }
}
vector<int> v[MAXN];
void dfs(int x) {
    for(int i = ; i < v[x].size(); i++) {
        dfs(v[x][i]);
        siz[x] += siz[v[x][i]];
    }
    f[len[x]] = max(f[len[x]], siz[x]);
}
int main() {
#ifdef WIN32
    freopen("a.in", "r", stdin);
#else
    //freopen("pow.in", "r", stdin);
    //freopen("pow.out", "w", stdout);
#endif
    scanf("%s", s + );
    N = strlen(s + );
    for(int i = ; i <= N; i++) insert(s[i] - 'a');
    for(int i = ; i <= tot; i++) v[fa[i]].push_back(i);
    dfs(root);
    for(int i = ; i <= N; i++)
        printf("%d\n", f[i]);
    return ;
}