bzoj3160 万径人踪灭

题意：给定只有ab的字符串，求其中不连续非空回文子序列的个数。

解：用所有回文子序列减去回文子串。

容易想到枚举中线。

设f[i]表示以i为中线，回文字符的个数。那么回文子序列就是∑(2^f[i] - 1)

怎么求f[i]呢？卷积！

我们考虑把i变成中线 * 2，那么f[i] = ( ∑(s[j] == s[i - j])  + 1) / 2

令a = 1，b = 0，那么卷积得出的就是所有a的回文字符数。同理可以求得所有b的回文字符数。

然后用回文自动机求一遍回文子串的数目，相减即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

typedef long long LL;
const int N = 100010;
const LL MO = 1e9 + 7;
const double pi = 3.1415926535897932384626;

struct cp {
    double x, y;
    cp(double X = 0, double Y = 0) {
        x = X;
        y = Y;
    }
    inline cp operator +(const cp &w) const {
        return cp(x + w.x, y + w.y);
    }
    inline cp operator -(const cp &w) const {
        return cp(x - w.x, y - w.y);
    }
    inline cp operator *(const cp &w) const {
        return cp(x * w.x - y * w.y, x * w.y + y * w.x);
    }
}a[N << 2];

int r[N << 2], f[N << 1];
char s[N];

inline LL qpow(LL a, int b) {
    LL ans = 1;
    while(b) {
        if(b & 1) {
            ans = ans * a % MO;
        }
        a = a * a % MO;
        b = b >> 1;
    }
    return ans;
}

inline void FFT(int n, cp *a, int f) {
    for(int i = 0; i < n; i++) {
        if(i < r[i]) {
            std::swap(a[i], a[r[i]]);
        }
    }

    for(int len = 1; len < n; len <<= 1) {
        cp Wn(cos(pi / len), f * sin(pi / len));
        for(int i = 0; i < n; i += (len << 1)) {
            cp w(1, 0);
            for(int j = 0; j < len; j++) {
                cp t = a[i + len + j] * w;
                a[i + len + j] = a[i + j] - t;
                a[i + j] = a[i + j] + t;
                w = w * Wn;
            }
        }
    }

    if(f == -1) {
        for(int i = 0; i <= n; i++) {
            a[i].x /= n;
        }
    }
    return;
}

namespace pam {
    int tr[N][26], fail[N], cnt[N], len[N], last, tot;
    inline void init() {
        len[1] = -1;
        fail[0] = fail[1] = 1;
        tot = last = 1;
        return;
    }
    inline int getfail(int d, int x) {
        while(s[d - len[x] - 1] != s[d]) {
            x = fail[x];
        }
        return x;
    }
    inline void insert(int d) {
        int f = s[d] - 'a';
        int p = getfail(d, last);
        if(!tr[p][f]) {
            ++tot;
            len[tot] = len[p] + 2;
            fail[tot] = tr[getfail(d, fail[p])][f];
            tr[p][f] = tot;
        }
        last = tr[p][f];
        cnt[last]++;
        return;
    }
    inline LL count() {
        LL ans = 0;
        for(int i = tot; i >= 2; i--) {
            ans = (ans + cnt[i]) % MO;
            (cnt[fail[i]] += cnt[i]) %= MO;
        }
        return ans;
    }
}

int main() {
    scanf("%s", s);
    int n = strlen(s) - 1;
    for(int i = 0; i <= n; i++) {
        a[i].x = (s[i] == 'a');
    }
    int len = 2, lm = 1;
    while(len <= n + n) {
        len <<= 1;
        lm++;
    }
    for(int i = 1; i <= len; i++) {
        r[i] = (r[i >> 1] >> 1) | ((i & 1) << (lm - 1));
    }

    FFT(len, a, 1);
    for(int i = 0; i <= len; i++) {
        a[i] = a[i] * a[i];
    }
    FFT(len, a, -1);
    for(int i = 0; i <= n + n; i++) {
        f[i] = ((int)(a[i].x + 0.5) + 1) >> 1;
    }

    for(int i = 0; i <= n; i++) {
        a[i].x = (s[i] == 'b');
        a[i].y = 0;
    }
    for(int i = n + 1; i <= len; i++) {
        a[i] = cp(0, 0);
    }
    FFT(len, a, 1);
    for(int i = 0; i <= len; i++) {
        a[i] = a[i] * a[i];
    }
    FFT(len, a, -1);
    for(int i = 0; i <= n + n; i++) {
        f[i] += ((int)(a[i].x + 0.5) + 1) >> 1;
    }

    LL ans = 0;
    for(int i = 0; i <= n + n; i++) {
        ans = (ans + qpow(2ll, f[i]) - 1) % MO;
    }
    pam::init();
    for(int i = 0; i <= n; i++) {
        pam::insert(i);
    }
    ans = (ans - pam::count() + MO) % MO;
    printf("%lld", ans);
    return 0;
}