【洛谷P4173】残缺的字符串

题目大意：给定一个文本串和一个模板串，串中含有通配符，求文本串中有多少个位置可以与文本串完全匹配。

题解：利用卷积求解字符串匹配问题。
通配符字符串匹配的数值表示为 $$\sum\limits_{i = 0}^{m - 1}(a[i] - b[i + k])^2 a[i]b[i + k]=0$$。直接展开之后计算三个卷积即可。
需要注意的是：并不是所有 a[i] b[i + k] 均为循环卷积，是否需要倍增取决于是否成环。

代码如下

#include <bits/stdc++.h>
using namespace std;
typedef complex<double> cp;
const double eps = 1e-8;
const double pi = acos(-1);

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, m;
	cin >> m >> n;
	string s, t;
	cin >> t >> s;
	int tot = 1, bit = 0;
	while (tot <= n + m) {
		tot <<= 1;
		++bit;
	}
	vector<int> rev(tot);
	for (int i = 0; i < tot; i++) {
		rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1;
	}
	vector<cp> a1(tot), a2(tot), a3(tot), b1(tot), b2(tot), b3(tot);
	auto work = [](char ch) {
		if (ch == '*') {
			return 0;
		} else {
			return ch - 'a' + 1;
		}
	};
	for (int i = 0; i < m; i++) {
		int x = work(t[i]);
		a1[m - 1 - i] = x;
		a2[m - 1 - i] = x * x;
		a3[m - 1 - i] = x * x * x;
	}
	for (int i = 0; i < n; i++) {
		int x = work(s[i]);
		b1[i] = x;
		b2[i] = x * x;
		b3[i] = x * x * x;
	}
	auto fft = [=](vector<cp> &v, int opt) {
		for (int i = 0; i < tot; i++) {
			if (i < rev[i]) {
				swap(v[i], v[rev[i]]);
			}
		}
		for (int mid = 1; mid < tot; mid <<= 1) {
			cp wn(cos(pi / mid), opt * sin(pi / mid));
			for (int j = 0; j < tot; j += mid << 1) {
				cp w(1, 0);
				for (int k = 0; k < mid; k++) {
					cp x = v[j + k], y = w * v[j + mid + k];
					v[j + k] = x + y, v[j + mid + k] = x - y;
					w *= wn;
				}
			}
		}
		if (opt == -1) {
			for (int i = 0; i < tot; i++) {
				v[i].real(round(v[i].real() / tot));
			}
		}
	};
	auto calc = [=](vector<cp> &a, vector<cp> &b) {
		fft(a, 1), fft(b, 1);
		for (int i = 0; i < tot; i++) {
			a[i] *= b[i];
		}
		fft(a, -1);
	};
	calc(a3, b1), calc(a2, b2), calc(a1, b3);
	int ans = 0;
	vector<int> pos;
	for (int k = 0; k < n - m + 1; k++) {
		double ret = a3[m + k - 1].real() - 2 * a2[m + k - 1].real() + a1[m + k - 1].real();
		if (fabs(ret) < eps) {
			++ans;
			pos.push_back(k + 1);
		}
	}
	cout << ans << endl;
	for (auto p : pos) {
		cout << p << " ";
	}
	return 0;
}