spoj Substrings【后缀自动机】

题目大意：求一个串里面长度分别为 1 ~ n 的子串中出现次数最多的次数

考虑每个点代表的字符串出现的次数为它right集合的大小，right集合大小可以遍历子树得到，用每个点的right集合大小去更新他 mx 的答案

#include 
#include 
#include 
#include 
#define N 500005
#define INF 1000000000
using namespace std;

int n,m,last = 1,tot = 1,p,q,np,nq;
int son[N][26],par[N],mx[N];
int ans[N],f[N],P[N],sm[N];
char s[N];

int new_node(int x)
{
    mx[++ tot] = x;
    return tot;
}

void add(int x)
{
    p = last;
    np = new_node(mx[p] + 1);
    for (;p && !son[p][x];p = par[p]) son[p][x] = np;
    if (!p) par[np] = 1;
    else
    {
        q = son[p][x];
        if (mx[q] == mx[p] + 1) par[np] = q;
        else
        {
            nq = new_node(mx[p] + 1);
            memcpy(son[nq],son[q],sizeof(son[nq]));
            par[nq] = par[q],par[q] = par[np] = nq;
            for (;son[p][x] == q;p = par[p]) son[p][x] = nq;
        }
    }
    f[last = np] ++;
}

int main()
{
    scanf("%s",s + 1),n = strlen(s + 1);
    for (int i = 1;i <= n;i ++) add(s[i] - 'a');
    for (int i = 1;i <= tot;i ++) sm[mx[i]] ++;
    for (int i = 1;i <= n;i ++) sm[i] += sm[i - 1];
    for (int i = 1;i <= tot;i ++) P[sm[mx[i]] --] = i;
    for (int i = tot;i > 1;i --) f[par[P[i]]] += f[P[i]];
    for (int i = 1;i <= tot;i ++) ans[mx[i]] = max(ans[mx[i]],f[i]);
    for (int i = n;i;i --) f[i] = max(f[i],f[i + 1]);
    for (int i = 1;i <= n;i ++) printf("%d\n",ans[i]);

    return 0;
}