APIO2014 回文串 回文自动机

APIO2014 回文串 回文自动机

题意

定义\(s\)的一个子串的存在值为这个子串出现的次数乘子串的长度

求\(s\)的所有回文串的存在值

\[1 \leq |s| \leq 300000 \]

分析

对\(s\)构建出回文自动机，\(cnt\)表示当前结点当前的出现次数，那么类似AC自动机的fail树，倒序对fail树的结点求siz即可得到结点的总出现次数

然后乘上len即可

代码

#include<bits/stdc++.h>
#define pii pair<int,int>
#define fi first
#define se second
#define eps 1e-9
#define db long double
#define equals(a,b) fabs(a-b) < eps
using namespace std;

typedef long long ll;



inline ll rd(){
	ll x;
	scanf("%lld",&x);
	return x;
}

const int maxn = 3e5 + 5;
char ss[maxn];

namespace pam {
	int sz, tot, last;
	int cnt[maxn], ch[maxn][26], len[maxn], fail[maxn];
	char s[maxn];
	int node(int l) {
  		sz++;
  		memset(ch[sz], 0, sizeof(ch[sz]));
  		len[sz] = l;
  		fail[sz] = cnt[sz] = 0;
  		return sz;
	}
	void clear() {
  		sz = -1;
  		last = 0;
  		s[tot = 0] = '$';
  		node(0);
  		node(-1);
  		fail[0] = 1;
	}
	int getfail(int x) {
  		while (s[tot - len[x] - 1] != s[tot]) x = fail[x];
  		return x;
	}
	void insert(char c) {
  		s[++tot] = c;
  		int now = getfail(last);
  		if (!ch[now][c - 'a']) {
    		int x = node(len[now] + 2);
    		fail[x] = ch[getfail(fail[now])][c - 'a'];
  	 	 	ch[now][c - 'a'] = x;
  		}
  		last = ch[now][c - 'a'];
  		cnt[last]++;
	}
}

int main(){
	scanf("%s",ss + 1);
	int len = strlen(ss + 1);
	pam::clear();
	for(int i = 1;i <= len;i++)
		pam::insert(ss[i]);
	ll ans = 0;
	for(int i = pam::sz;i >= 1;i--)
		pam::cnt[pam::fail[i]] += pam::cnt[i];
	for(int i = 1;i <= pam::sz;i++)
		ans = max(ans,(ll)(pam::cnt[i]) * (pam::len[i]));
	cout << ans;
}