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FFT字符串匹配（解决通配符问题）
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FFT字符串匹配
定義字符串下標從000，開始，有文本串AAA長度為nnn，模式串BBB長度為mmm，我們可以考慮一個函數f(x,y)=A(x)?B(y)f(x, y) = A(x) - B(y)f(x,y)=A(x)?B(y)。
我們設F(x)(x≥m?1)=∑i=0m?1f(x?m+1+i,i)F(x)(x \ge m - 1) = \sum\limits_{i = 0} ^{m - 1} f(x - m + 1 + i, i)F(x)(x≥m?1)=i=0∑m?1?f(x?m+1+i,i)，由定義顯然可以得到如果F(x)=0F(x) = 0F(x)=0，則A[x?m+1,x]=BA[x - m + 1, x] = BA[x?m+1,x]=B也就是兩個字符匹配上了，
但是考慮"ab","ba""ab", "ba""ab","ba"兩個字符串，他們也是匹配的，我們稍微修改一下f(x,y)f(x, y)f(x,y)函數令其為：(A(x)?B(y))2(A(x) - B(y)) ^ 2(A(x)?B(y))2，這樣這個函數就沒有問題了。
我們考慮翻轉一下BBB串，令其為SSS，則有B(i)=S(m?i?1)B(i) = S(m - i - 1)B(i)=S(m?i?1)，
F(x)=∑i=0m?1(A(x?m+1+i)?S(m?i?1))2F(x)=∑i=0m?1S(m?i?1)2+∑i=0m?1A(x?m+1+i)2?2×∑i=0m?1A(x?m+1+i)S(m?i?1)F(x) = \sum\limits_{i = 0} ^{m - 1} \left(A(x - m + 1 + i) - S(m - i - 1)\right) ^ 2\\ F(x) = \sum_{i = 0} ^{m - 1} S(m - i - 1) ^ 2 + \sum_{i = 0} ^{m - 1} A(x - m + 1 + i) ^ 2 - 2 \times \sum_{i = 0} ^{m - 1} A(x - m + 1 + i) S(m - i - 1)\\ F(x)=i=0∑m?1?(A(x?m+1+i)?S(m?i?1))2F(x)=i=0∑m?1?S(m?i?1)2+i=0∑m?1?A(x?m+1+i)2?2×i=0∑m?1?A(x?m+1+i)S(m?i?1)
第一項是一個定值，第二可以O(n)O(n)O(n)預處理，然后前綴和O(1)O(1)O(1)得到，第三項不難發現是一個卷積的形式，所以可以通過FFTFFTFFT得到，整體復雜度O(nlog?n)O(n \log n)O(nlogn)。
以上我們已經可以解決當模式串的字符串匹配了，盡管復雜度不如KMPKMPKMP優秀，但是我們考慮一個缺項字符串匹配：
a*b
aebr*ob
我們考慮重新設計f(x,y)f(x, y)f(x,y)函數，定義f(x,y)=(A(x)?B(y))2A(x)B(y)f(x, y) = (A(x) - B(y)) ^ 2 A(x) B(y)f(x,y)=(A(x)?B(y))2A(x)B(y)，同樣的考慮翻轉BBB串，有B(i)=S(m?1?i)B(i) = S(m - 1 - i)B(i)=S(m?1?i)。
F(x)=∑i=0m?1(A(x?m+1+i)?S(m?1?i))2A(x?m+1+i)S(m?1?i)∑i=0m?1A(x?m+1+i)S(m?1?i)3+∑i=0m?1A(x?m+1+i)3S(m?1?i)?2×∑i=0m?1A(x?m+1+i)2S(m?1?i)2F(x) = \sum_{i = 0} ^{m - 1} (A(x - m + 1 + i) - S(m - 1 - i)) ^ 2 A(x - m + 1 + i) S(m - 1 - i)\\ \sum_{i = 0} ^{m - 1}A(x - m + 1 + i) S(m - 1 - i) ^ 3 + \sum_{i = 0} ^{m - 1} A(x - m + 1 + i) ^ 3 S(m - 1 - i) - 2 \times \sum_{i = 0} ^{m - 1} A(x - m + 1 + i) ^ 2 S(m - 1 - i) ^ 2\\ F(x)=i=0∑m?1?(A(x?m+1+i)?S(m?1?i))2A(x?m+1+i)S(m?1?i)i=0∑m?1?A(x?m+1+i)S(m?1?i)3+i=0∑m?1?A(x?m+1+i)3S(m?1?i)?2×i=0∑m?1?A(x?m+1+i)2S(m?1?i)2
容易發現這里是三個多項式相加的形式，所以只要做三次FFTFFTFFT即可得到答案，放一個模板題。
#include <bits/stdc++.h>using namespace std
;struct Complex {double r
, i
;Complex(double _r
= 0, double _i
= 0) : r(_r
), i(_i
) {}
};Complex
operator + (const Complex
&a
, const Complex
&b
) {return Complex(a
.r
+ b
.r
, a
.i
+ b
.i
);
}Complex
operator - (const Complex
&a
, const Complex
&b
) {return Complex(a
.r
- b
.r
, a
.i
- b
.i
);
}Complex
operator * (const Complex
&a
, const Complex
&b
) {return Complex(a
.r
* b
.r
- a
.i
* b
.i
, a
.r
* b
.i
+ a
.i
* b
.r
);
}Complex
operator / (const Complex
&a
, const Complex
&b
) {return Complex((a
.r
* b
.r
+ a
.i
* b
.i
) / (b
.r
* b
.r
+ b
.i
* b
.i
), (a
.i
* b
.r
- a
.r
* b
.i
) / (b
.r
* b
.r
+ b
.i
* b
.i
));
}Complex
operator * (const Complex
&a
, const double &b
) {return Complex(a
.r
* b
, a
.i
* b
);
}const int N
= 2e6 + 10;int r
[N
];void get_r(int lim
) {for (int i
= 0; i
< lim
; i
++) {r
[i
] = (i
& 1) * (lim
>> 1) + (r
[i
>> 1] >> 1);}
}void FFT(Complex
*f
, int lim
, int rev
) {for (int i
= 0; i
< lim
; i
++) {if (i
< r
[i
]) {swap(f
[i
], f
[r
[i
]]);}}const double pi
= acos(-1.0);for (int mid
= 1; mid
< lim
; mid
<<= 1) {Complex wn
= Complex(cos(pi
/ mid
), rev
* sin(pi
/ mid
));for (int len
= mid
<< 1, cur
= 0; cur
< lim
; cur
+= len
) {Complex w
= Complex(1, 0);for (int k
= 0; k
< mid
; k
++, w
= w
* wn
) {Complex x
= f
[cur
+ k
], y
= w
* f
[cur
+ mid
+ k
];f
[cur
+ k
] = x
+ y
, f
[cur
+ mid
+ k
] = x
- y
;}}}if (rev
== -1) {for (int i
= 0; i
< lim
; i
++) {f
[i
].r
/= lim
;}}
}Complex a
[N
], b
[N
], c
[N
];char str1
[N
], str2
[N
];int A
[N
], S
[N
], n
, m
, lim
;int main() {scanf("%d %d %s %s", &m
, &n
, str2
, str1
);for (int i
= 0; i
< n
; i
++) {A
[i
] = str1
[i
] == '*' ? 0 : str1
[i
] - 'a' + 1;}for (int i
= 0; i
< m
; i
++) {S
[i
] = str2
[m
- i
- 1] == '*' ? 0 : str2
[m
- i
- 1] - 'a' + 1;}lim
= 1;while (lim
< n
+ m
) {lim
<<= 1;}get_r(lim
);for (int i
= 0; i
< lim
; i
++) {b
[i
] = Complex(A
[i
], 0);c
[i
] = Complex(S
[i
] * S
[i
] * S
[i
], 0);}FFT(b
, lim
, 1), FFT(c
, lim
, 1);for (int i
= 0; i
< lim
; i
++) {a
[i
] = a
[i
] + b
[i
] * c
[i
];}for (int i
= 0; i
< lim
; i
++) {b
[i
] = Complex(A
[i
] * A
[i
] * A
[i
], 0);c
[i
] = Complex(S
[i
], 0);}FFT(b
, lim
, 1), FFT(c
, lim
, 1);for (int i
= 0; i
< lim
; i
++) {a
[i
] = a
[i
] + b
[i
] * c
[i
];}for (int i
= 0; i
< lim
; i
++) {b
[i
] = Complex(A
[i
] * A
[i
], 0);c
[i
] = Complex(S
[i
] * S
[i
], 0);}FFT(b
, lim
, 1), FFT(c
, lim
, 1);for (int i
= 0; i
< lim
; i
++) {a
[i
] = a
[i
] - 2 * b
[i
] * c
[i
];}FFT(a
, lim
, -1);vector
<int> ans
;for (int i
= m
- 1; i
< n
; i
++) {if ((long long)(a
[i
].r
+ 0.5) == 0) {ans
.push_back(i
- m
+ 2);}}printf("%d\n", ans
.size());for (auto it
: ans
) {printf("%d ", it
);}puts("");return 0;
}
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