Luogu P4173 残缺的字符串-FFT在字符串匹配中的应用

P4173 残缺的字符串

FFT在字符串匹配中的应用.

能解决大概这种问题:

给定长度为\(m\)的A串,长度为\(n\)的B串。问A串在B串中的匹配数

我们设一个函数(下标从\(0\)开始)

\(C(x,y) =A(x)- B(y)\)，若为0,表示B串中以第\(y\)个字符结尾的字符可以与A串中以\(x\)节为结尾的字符可以匹配

\(P(x) = \sum_{i = 0}^{m - 1}C(i,x - m + i + 1)\)

但是很遗憾当\(P(x)\),等于零时,只能够说明上述子串的字符集相同.

为什么？因为负数的存在！

我们考虑怎么去掉负数,平方！

\(P(x) = \sum_{i = 0}^{m - 1}(A(i) - B[x - m + i + 1])^2\)

这时候,如果上式为\(0\)，就能证明B串中\(x\)结尾的串可以与A匹配.

老样子设

\(f(i) = A(m - i - 1)\)

\(g(i) = B(i)\)

则有

\(P(x) = \sum_{i = 0}^{m - 1}f(m - i - 1)^2 -\sum_{i = 0}^{m - 1}2f(m - i - 1)g(x - m + i + 1) -\sum_{i = 0}^{m - 1}g(x - m + i + 1)^2\)

发现第一项和第三项是可以通过处理前缀和搞出来的！

而第二项是个卷积,我们只需要求\(P(x)\)是否为零就好了。

我们终于这到了题目上.

这道题目中含有通配符,上式很明显不再成立

但大体思路还是不变的

\(C(x)\)与\(P(x)\)的意义不变

我们设

\(P(x) = \sum_{i = 0}^{m - 1}(A(i) - B(x - m + i + 1))^2A(i)B(x - m + i + 1)\)

即当B串\(x\)的位置为通配符时,\(B(x) = 0\)，A同理

这样我们就又能用\(P(x)\)表示能否匹配了

同理，设\(f(x)\)与\(g(x)\)意义同上

\(P(x) =\sum_{i = 0}^{m - 1}f(m - i - 1)^3g(x - m + i + 1) - \sum_{i = 0}^{m - 1}f(m - i - 1)^2g(x - m + i + 1)+\sum_{i = 0}^{m - 1}f(m - i - 1)g(x - m + i + 1)^3\)

然后发现

上式三项都是卷积！

所以我们跑7遍FFT就好了

#include<cstdio>
#include<iostream>
#include<cctype>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
const int N = 3e5 + 3;
const double Pi = acos(-1.0);
const double eps = 1e-12;
struct point{
    double x,y;
    point(double xx = 0,long double yy = 0){
        x = xx,y = yy;
    }
}a[N << 2],b[N << 2],c[N << 2];
char s1[N],s2[N];
int c1[N],c2[N];
int r[N << 2];
int n,m,limit = 1,l;
vector <int> G;
point operator + (point a,point b){return point(a.x + b.x,a.y + b.y);}
point operator - (point a,point b){return point(a.x - b.x,a.y - b.y);}
point operator * (point a,point b){return point(a.x * b.x - a.y * b.y,a.x * b.y + a.y * b.x);}
inline void fftle(point *A,int type){
    for(int i = 0;i < limit;++i)
        if(i < r[i]) swap(A[i],A[r[i]]);
    for(int mid = 1;mid < limit;mid <<= 1){
        point Wn = point(cos(Pi / mid),type * sin(Pi / mid));
        for(int R = mid << 1,j = 0;j < limit;j += R){
            point w(1,0);
            for(int k = 0;k < mid;++k,w = w * Wn){
                point x = A[j + k],y = A[j + mid + k] * w;
                A[j + k] = x + y;
                A[j + mid + k] = x - y;
            }
        }
    }
    if(type == -1) for(int i = 0;i < limit;++i) A[i].x = A[i].x / limit;
}
int main(){
    scanf("%d%d",&m,&n);
    scanf("%s%s",s1,s2);
    point zero = point(0,0);
    for(int i = 0;i < m;++i) c1[i] = s1[m - i - 1] == '*' ? 0 : s1[m - i - 1] - 'a' + 1;
    for(int i = 0;i < n;++i) c2[i] = s2[i] == '*' ? 0 : s2[i] - 'a' + 1;
    while(limit <= (n + m)) limit <<= 1,l++;
    for(int i = 0;i < limit;++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
    for(int i = 0;i < m;++i) a[i].x = c1[i] * c1[i] * c1[i];
    for(int i = 0;i < n;++i) b[i].x = c2[i];
    fftle(a,1);fftle(b,1);
    for(int i = 0;i < limit;++i) c[i] = c[i] + (a[i] * b[i]),a[i] = b[i] = zero;
    for(int i = 0;i < m;++i) a[i].x = c1[i] * c1[i];
    for(int i = 0;i < n;++i) b[i].x = c2[i] * c2[i];
    fftle(a,1);fftle(b,1);
    point w(2,0);
    for(int i = 0;i < limit;++i) c[i] = c[i] - ((a[i] * b[i]) * w),a[i] = b[i] = zero;
    for(int i = 0;i < m;++i) a[i].x = c1[i];
    for(int i = 0;i < n;++i) b[i].x = c2[i] * c2[i] * c2[i];
    fftle(a,1);fftle(b,1);
    for(int i = 0;i < limit;++i) c[i] = c[i] + (a[i] * b[i]);
    fftle(c,-1);
    //for(int i = m - 1;i < n;++i) printf("%lf ",fabs(c[i].x / limit));puts("");
    for(int i = m - 1;i < n;++i) if((fabs)(c[i].x) < 0.5) G.push_back(i + 2 - m);
    printf("%d\n",(int)G.size());
    for(int i = 0;i < (int)G.size();++i) printf("%d ",G[i]);
}

参考博客