标签:sig put 字符 str 技巧 typedef int ever clu
【题目大意】
双串带通配符匹配。
$|S|, |T| \leq 5 * 10^5$
TL: 2s
【题解】
参考bzoj 4503
可以设计如下函数 A[i] * B[i] * (A[i] - B[i])^2
如果有通配符,A[i] = 0,否则,A[i] = s[i] - ‘a‘ + 1;B同理。
可以自行验证,这是一种很妙的设计。
然后就是卷积的事情了。大概做9次DFT。
可以用类似于MTT的技巧搞到4次,不会写。
# include <math.h> # include <stdio.h> # include <string.h> # include <iostream> # include <algorithm> // # include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; const int M = 5e5 + 10, N = 2e6 + 10; const int mod = 1e9+7; const ld pi = acos(-1.0); char s[M], t[M]; int A[M], B[M], ns, nt; struct cp { ld x, y; cp() {} cp(ld x, ld y) : x(x), y(y) {} friend cp operator + (cp a, cp b) { return cp(a.x + b.x, a.y + b.y); } friend cp operator - (cp a, cp b) { return cp(a.x - b.x, a.y - b.y); } friend cp operator * (cp a, cp b) { return cp(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); } }a[N], b[N], ans[N]; namespace FFT { int n, lst[N]; cp w[2][N]; inline void set(int _n) { n = 1; while(n < _n) n <<= 1; for (int i=0; i<n; ++i) w[0][i] = cp(cos(pi * 2 / n * i), sin(pi * 2 / n * i)), w[1][i] = cp(w[0][i].x, -w[0][i].y); int len = 0; while((1<<len) < n) ++len; for (int i=0; i<n; ++i) { int t = 0; for (int j=0; j<len; ++j) if(i & (1<<j)) t |= (1<<(len-j-1)); lst[i] = t; } } inline void DFT(cp *a, int op) { cp *o = w[op]; for (int i=0; i<n; ++i) if(i < lst[i]) swap(a[i], a[lst[i]]); for (int len=2; len<=n; len<<=1) { int m = len>>1; for (cp *p=a; p!=a+n; p+=len) { for (int k=0; k<m; ++k) { cp t = o[n/len*k] * p[k+m]; p[k+m] = p[k] - t; p[k] = p[k] + t; } } } if(op) { for (int i=0; i<n; ++i) a[i].x /= (ld)n, a[i].y /= (ld)n; } } } # define L FFT::n int main() { scanf("%s%s", t, s); ns = strlen(s), nt = strlen(t); for (int i=0; i<ns; ++i) A[i] = (s[i] == ‘*‘ ? 0 : s[i] - ‘a‘ + 1); for (int i=0; i<nt; ++i) B[i] = (t[i] == ‘*‘ ? 0 : t[i] - ‘a‘ + 1); reverse(B, B+nt); // (A[i] - B[i])^2 * A[i] * B[i] = A[i]^3 * B[i] + A[i] * B[i]^3 - A[i]^2 * B[i]^2 FFT :: set(max(ns, nt)); for (int i=0; i<ns; ++i) a[i] = cp(A[i] * A[i] * A[i], 0); for (int i=ns; i<L; ++i) a[i] = cp(0, 0); for (int i=0; i<nt; ++i) b[i] = cp(B[i], 0); for (int i=nt; i<L; ++i) b[i] = cp(0, 0); FFT :: DFT(a, 0); FFT :: DFT(b, 0); for (int i=0; i<L; ++i) ans[i] = ans[i] + a[i] * b[i]; for (int i=0; i<ns; ++i) a[i] = cp(A[i], 0); for (int i=ns; i<L; ++i) a[i] = cp(0, 0); for (int i=0; i<nt; ++i) b[i] = cp(B[i] * B[i] * B[i], 0); for (int i=nt; i<L; ++i) b[i] = cp(0, 0); FFT :: DFT(a, 0); FFT :: DFT(b, 0); for (int i=0; i<L; ++i) ans[i] = ans[i] + a[i] * b[i]; for (int i=0; i<ns; ++i) a[i] = cp(A[i] * A[i] * 2, 0); for (int i=ns; i<L; ++i) a[i] = cp(0, 0); for (int i=0; i<nt; ++i) b[i] = cp(B[i] * B[i], 0); for (int i=nt; i<L; ++i) b[i] = cp(0, 0); FFT :: DFT(a, 0); FFT :: DFT(b, 0); for (int i=0; i<L; ++i) ans[i] = ans[i] - a[i] * b[i]; FFT :: DFT(ans, 1); for (int i=nt-1; i<ns; ++i) { if((int)(ans[i].x-0.5) == 0) printf("%d ", i-nt+2); } puts(""); return 0; }
可能是noi前最后一次复习FFT了。
标签:sig put 字符 str 技巧 typedef int ever clu
原文地址:http://www.cnblogs.com/galaxies/p/20170707_14.html