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 <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define N 500005
#define INF 1000000000
using namespace std;

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

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;
}