luogu P4199 万径人踪灭

嘟嘟嘟


方案：回文子序列数 - 回文子串数。
回文子串数用manacher解决就行了，关键是怎么求会问序列数。
一个比较好的\(O(n ^ 2)\)的算法：对于一个回文中心\(i\)，\(O(n)\)求出以\(i\)为中心位置对称且字母相同的字母对数\(x\)，则以\(i\)为回文中心的回文子序列有\(2 ^ x - 1\)个（排除空序列）。
现在想一下怎么优化。
上面寻找的过程，用一个式子写出来就是这样：

\[c(i) = \sum _ {j = 0} ^ {i} s[i - j] == s[i + j] \]

会发现这东西和卷积特别像，但是怎么用多项式表示\(s[i - j] == s[i + j]\)呢？
因为题中说了只有\(a, b\)两种字母，所以构造两个只有\(0\)或\(1\)的多项式\(A, B\)，如果\(s[i] = 'a'\)，那么\(A(i) = 1\)；\(B\)同理。
这样的话式子就变成了

\[c(i) = \sum_{j = 0} ^ {i} A(i - j) * A(i + j) + \sum _ {j = 0} ^ {i} B(i - j) * B(i + j) - 1 \]

然后两次卷积就行啦。


fft还是不熟……

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const db PI = acos(-1);
const int maxn = 8e5 + 5;
const ll mod = 1e9 + 7;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, len = 1;
char s[maxn];

struct Comp
{
  db x, y;
  inline Comp operator + (const Comp& oth)const
  {
    return (Comp) {x + oth.x, y + oth.y};
  }
  inline Comp operator - (const Comp& oth)const
  {
    return (Comp){x - oth.x, y - oth.y};
  }
  inline Comp operator * (const Comp& oth)const
  {
    return (Comp){x * oth.x - y * oth.y, x * oth.y + oth.x * y};
  }
  friend inline void swap(Comp& a, Comp& b)
  {
    swap(a.x, b.x); swap(a.y, b.y);
  }
}a[maxn], b[maxn], omg[maxn], inv[maxn];

void init()
{
  omg[0] = inv[0] = (Comp){1, 0};
  omg[1] = inv[len - 1] = (Comp){cos(2 * PI / len), sin(2 * PI / len)};
  for(int i = 2; i < len; ++i) omg[i] = inv[len - i] = omg[i - 1] * omg[1];
}
void fft(Comp* a, Comp* omg)
{
  int lim = 0;
  while((1 << lim) < len) lim++;
  for(int i = 0; i < len; ++i)
    {
      int t = 0;
      for(int j = 0; j < lim; ++j) if((i >> j) & 1) t |= (1 << (lim - j - 1));
      if(i < t) swap(a[i], a[t]);
    }
  for(int l = 2; l <= len; l <<= 1)
    {
      int q = l >> 1;
      for(Comp* p = a; p != a + len; p += l)
	for(int i = 0; i < q; ++i)
	  {
	    Comp t = omg[len / l * i] * p[i + q];
	    p[i + q] = p[i] - t, p[i] = p[i] + t;
	  }
    }
}

char t[maxn << 1];
int p[maxn << 1], m;
void manacher()
{
  t[0] = '@';
  for(int i = 0; i < n; ++i) t[i << 1 | 1] = '#', t[(i << 1) + 2] = s[i];
  m = (n << 1) + 2;
  t[m - 1] = '#'; t[m] = '$'; 
  int mx = 0, id;
  for(int i = 1; i < m; ++i)
    {
      if(mx > i) p[i] = min(p[(id << 1) - i], mx - i);
      else p[i] = 1;
      while(t[i - p[i]] == t[i + p[i]]) p[i]++;
      if(i + p[i] > mx) mx = p[i] + i, id = i;
    }
}

ll quickpow(ll a, ll b)
{
  ll ret = 1;
  for(; b; b >>= 1, a = a * a % mod)
    if(b & 1) ret = ret * a % mod;
  return ret;
}

ll Ans = 0;
int ans[maxn];

int main()
{
  scanf("%s", s);
  n = strlen(s);
  while(len < (n << 1)) len <<= 1;
  for(int i = 0 ; i < n; ++i) a[i] = (Comp){s[i] == 'a', 0}, b[i] = (Comp){s[i] == 'b', 0};
  init();
  fft(a, omg); fft(b, omg);
  for(int i = 0; i < len; ++i) a[i] = a[i] * a[i] + b[i] * b[i];
  fft(a, inv);
  for(int i = 0; i < len; ++i) ans[i] = ((a[i].x / len + 0.5) + 1) / 2;
  for(int i = 0; i < len; ++i) Ans = (Ans + quickpow(2, ans[i]) - 1) % mod;
  manacher();
  for(int i = 1; i < m; ++i) Ans = (Ans - (p[i] >> 1) + mod) % mod;
  write(Ans), enter;
  return 0;
}