Bzoj3160:万径人踪灭

题面

Bzoj

Sol

求不连续回文子序列的个数
\(ans=\)回文子序列个数-连续回文子序列个数
即回文子序列个数-回文子串个数
后面直接\(Manacher\)就好了
考虑前面的
枚举对称轴,设\(f[i]\)表示对称轴\(i\)两边相同字符的对数
那么最终答案就是\(\sum 2^{f[i]}-1\)
考虑求\(f[i]\)
只有当原串中的两个字符相同才会有贡献
也就是\(s[i-x]=s[i+x]\)
单独考虑\(a\)\(b\)的贡献
\(f[i]=\sum [s[i-x]==s[i+x]]\)
设当前考虑\(a\)的贡献
把是\(a\)的设为\(1\)不是的为\(0\),做个卷积就可以求出\(f\)
那么就可以\(FFT\)
注意两个字符之间也算对称轴

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int Zsy(1e9 + 7); 
const int _(5e5 + 5);
const double PI(acos(-1));

int n, p[_];
ll f[_];
int N, M, l, r[_];
char s[_], a[_];
struct Complex{
    double real, image;
    IL Complex(){
		real = image = 0;
	}
	
    IL Complex(RG double a, RG double b){
		real = a, image = b;
	}
	
    IL Complex operator +(RG Complex B){
		return Complex(real + B.real, image + B.image);
	}
	
    IL Complex operator -(RG Complex B){
		return Complex(real - B.real, image - B.image);
	}
	
    IL Complex operator *(RG Complex B){
		return Complex(real * B.real - image * B.image, real * B.image + image * B.real);
		}
} A[_], B[_];

IL void FFT(RG Complex *P, RG int opt){
	for(RG int i = 0; i < N; ++i) if(i < r[i]) swap(P[i], P[r[i]]);
	for(RG int i = 1; i < N; i <<= 1){
		RG Complex W(cos(PI / i), opt * sin(PI / i));
		for(RG int j = 0, p = i << 1; j < N; j += p){
			RG Complex w(1, 0);
			for(RG int k = 0; k < i; ++k, w = w * W){
				RG Complex X = P[k + j], Y = w * P[k + j + i];
				P[k + j] = X + Y, P[k + j + i] = X - Y;
			}
		}
	}
}

IL void Mul(){
	FFT(A, 1);
	for(RG int i = 0; i < N; ++i) B[i] = A[i] * A[i];
	FFT(B, -1);
	for(RG int i = 0; i < N; ++i) B[i].real = B[i].real / N + 0.5;
	for(RG int i = 1; i <= M; ++i) f[i] += ((ll)(B[i].real) + 1) >> 1;
}

IL ll Manacher(){
	RG ll ans = 0; RG int mx = 0, len = 1; a[1] = '#';
	for(RG int i = 1; i <= n; ++i) a[++len] = s[i], a[++len] = '#';
	for(RG int i = 1, id = 0, mx = 0; i <= len; ++i){
		if(i < mx) p[i] = min(mx - i, p[(id << 1) - i]);
		while(i - p[i] && i + p[i] <= len && a[i - p[i]] == a[i + p[i]]) ++p[i];
		if(p[i] + i > mx) mx = p[i] + i, id = i;
		(ans += p[i] >> 1) %= Zsy;
	}
	return ans;
}

IL ll Pow(RG ll x, RG ll y){
	RG ll ret = 1;
	for(; y; y >>= 1, x = x * x % Zsy)
		if(y & 1) ret = ret * x % Zsy;
	return ret;
}

int main(RG int argc, RG char* argv[]){
	scanf(" %s", s + 1); n = strlen(s + 1);
	for(M = n + n, N = 1; N <= M; N <<= 1) ++l;
	for(RG int i = 0; i < N; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
	RG ll ans = -Manacher();
	for(RG int i = 0; i < N; ++i) A[i] = Complex(s[i] == 'a', 0);
	Mul();
	for(RG int i = 0; i < N; ++i) A[i] = Complex(s[i] == 'b', 0);
	Mul();
	for(RG int i = 1; i <= M; ++i) (ans += Pow(2, f[i]) - 1) % Zsy;
	printf("%lld\n", (ans + Zsy) % Zsy);
	return 0;
}
posted @ 2018-02-12 22:29  Cyhlnj  阅读(119)  评论(0编辑  收藏  举报