2019牛客暑期多校训练营（第四场） I-string 广义后缀自动机+回文自动机

题意

分析

考虑两个串如果反过来不相同个数怎么统计
其实就是正串和反串都放在一个广义后缀自动机上统计，然后 答案/2
这样对吗？我们考虑自己本身回文的情况，还要加上本质不同的回文串个数
直接上回文自动机就可以了

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 800010;
char s[N];
struct Sam
{
  ll fail[N],step[N],g[N][27],last,tot;
  void init()
  {
	memset(fail,0,sizeof(fail));
	memset(step,0,sizeof(step));
	memset(g,0,sizeof(g));
	last = tot = 1;
  }
  void ins(ll nx)
  {
	if(g[last][nx])
    {
	  ll p = last; ll q = g[last][nx];
	  if(step[q] == step[p] +1) last = q;
	  else
	  {
		ll nq = ++tot; step[nq] = step[p] + 1;
		fail[nq] = fail[q]; fail[q] = nq;
		for(ll i=1;i<=26;i++) g[nq][i] = g[q][i];
		while(g[p][nx] == q && p){g[p][nx] = nq,p=fail[p];} last = nq;
	  }return ;
	}
	ll p = last; ll np = ++tot; step[np] = step[p] + 1;
	while(!g[p][nx] && p) g[p][nx] = np,p=fail[p];
	if(!p) fail[np] = 1;
	else
	{
	  ll q = g[p][nx];
	  if(step[q] == step[p]+1){fail[np] = q;}
	  else
	  {
		ll nq = ++tot; step[nq] = step[p] + 1;
		fail[nq] = fail[q]; fail[np] = fail[q] = nq;
	    for(ll i=1;i<=26;i++) g[nq][i] = g[q][i];
		while(g[p][nx] == q && p) g[p][nx] = nq,p=fail[p];
	  }
	}last=np; 
  }
}SAM;

struct Pam
{
  ll fail[N],step[N],g[N][27],last,n,tot;
  void init()
  {
	memset(step,0,sizeof(step));
	memset(g,0,sizeof(g));
	fail[0] = fail[1] = 1; step[1] = -1; tot = 1;
  }
  void ins(ll nx)
  {
	ll p = last; n++;
	while(s[n] != s[n-step[p]-1]) p = fail[p];
	if(!g[p][nx])
	{
	  ll np = ++tot; step[np] = step[p] + 2;
	  ll q = fail[p];
	  while(s[n] != s[n-step[q]-1]) q = fail[q];
	  fail[np] = g[q][nx]; g[p][nx] = np; 
	}last = g[p][nx];
  }
}PAM;

int main()
{
  scanf("%s",s+1); ll len=strlen(s+1); SAM.init(); PAM.init();
  for(ll i=1;i<=len;i++)
  {
	SAM.ins(s[i]-'a'+1);
	PAM.ins(s[i]-'a'+1);
  }
  SAM.last = 1; for(ll i=len;i>=1;i--) SAM.ins(s[i]-'a'+1);
  ll ans = PAM.tot-1;
  for(ll i=1;i<=SAM.tot;i++) ans += SAM.step[i] - SAM.step[SAM.fail[i]];
  return printf("%lld\n",ans>>1),0;
}