将第二个字符串改成能赢对方时对方的字符并倒序后,字符串匹配就是卷积的过程。
那么就枚举字符做三次卷积即可。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> struct Complex { double r, i; Complex(){} Complex(double r, double i): r(r), i(i) {} Complex operator + (const Complex &p) const { return Complex(r + p.r, i + p.i); } Complex operator - (const Complex &p) const { return Complex(r - p.r, i - p.i); } Complex operator * (const Complex &p) const { return Complex(r * p.r - i * p.i, r * p.i + i * p.r); } }; const int N = 4e5 + 7; const double pi = acos(-1.0); int r[N]; void FFT(Complex a[], int n, int pd) { for (int i = 0; i < n; i++) if (i < r[i]) std::swap(a[i], a[r[i]]); for (int mid = 1; mid < n; mid <<= 1) { Complex wn(cos(pi / mid), pd * sin(pi / mid)); for (int l = mid << 1, j = 0; j < n; j += l) { Complex w(1.0, 0.0); for (int k = 0; k < mid; k++, w = w * wn) { Complex u = a[k + j], v = w * a[k + j + mid]; a[k + j] = u + v; a[k + j + mid] = u - v; } } } if (pd == -1) for (int i = 0; i < n; i++) a[i] = Complex(a[i].r / n, a[i].i / n); } Complex A[N], B[N]; int n, m, limit, sum[N]; char s[N], t[N]; void solve(char ch) { for (int i = 0; i < n; i++) A[i] = Complex(s[i] == ch ? 1 : 0, 0.0); for (int i = n; i < limit; i++) A[i] = Complex(0.0, 0.0); for (int i = 0; i < m; i++) B[i] = Complex(t[i] == ch ? 1.0 : 0.0, 0.0); for (int i = m; i < limit; i++) B[i] = Complex(0.0, 0.0); FFT(A, limit, 1); FFT(B, limit, 1); for (int i = 0; i < limit; i++) A[i] = A[i] * B[i]; FFT(A, limit, -1); for (int i = 0; i < limit; i++) sum[i] += (int)(A[i].r + 0.5); } int main() { scanf("%d%d", &n, &m); scanf("%s", s); scanf("%s", t); for (int i = 0; i < m; i++) { if (t[i] == 'R') t[i] = 'S'; else if (t[i] == 'S') t[i] = 'P'; else t[i] = 'R'; } std::reverse(t, t + m); int len = n + m; int l = 0; limit = 1; while (limit < len) limit <<= 1, l++; for (int i = 0; i < limit; i++) r[i] = r[i >> 1] >> 1 | ((i & 1) << (l - 1)); solve('R'); solve('S'); solve('P'); int ans = 0; for (int i = m - 1; i < limit; i++) ans = std::max(ans, sum[i]); printf("%d\n", ans); return 0; }