并不对劲的复健训练-有通配符的字符串匹配

最新推荐文章于 2024-07-22 00:45:33 发布

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最新推荐文章于 2024-07-22 00:45:33 发布

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原文链接：http://www.cnblogs.com/xzyf/p/11598961.html

本文深入讲解了一种基于字符串匹配的算法实现，通过将匹配问题转化为数学表达式的值为零的问题，利用复杂数学运算和数据结构进行高效匹配。文章详细介绍了算法的实现过程，包括数据读取、写入、快速傅立叶变换等关键技术，并提供了完整的代码示例。

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这个人讲得很清楚

关键在于把“匹配”转换成“某个式子的值为0”。

#include<algorithm>
#include<cmath>
#include<complex>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
#define view(u,k) for(int k=fir[u];~k;k=nxt[k])
#define LL long long
#define cd complex<double>
#define maxn 2400007
using namespace std;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)&&ch!='-')ch=getchar();
    if(ch=='-')f=-1,ch=getchar();
    while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
    return x*f;
}
void write(int x)
{
    if(x==0){putchar('0'),putchar(' ');return;}
    int f=0;char ch[20];
    if(x<0)putchar('-'),x=-x;
    while(x)ch[++f]=x%10+'0',x/=10;
    while(f)putchar(ch[f--]);
    putchar(' ');
    return;
}
const double pi=acos(-1);
int l1,l2,r[maxn],len,nown,ans[maxn],num;
cd a[maxn],b[maxn],A[maxn],B[maxn],c[maxn];
char s1[maxn],s2[maxn];
void getlen(int li)
{
    for(len=0,nown=1;nown<li;nown<<=1,len++);
    rep(i,0,nown-1)r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
}
void dft(cd * u,double f)
{
    rep(i,0,nown-1)if(i<r[i])swap(u[i],u[r[i]]);
    for(int i=1;i<nown;i<<=1)
    {
        cd wn(cos(pi/i),sin(f*pi/i)),x,y;
        for(int j=0;j<nown;j+=(i<<1))
        {
            cd w(1,0);
            rep(k,0,i-1)
                x=u[j+k],y=w*u[i+j+k],u[j+k]=x+y,u[i+j+k]=x-y,w*=wn;
        }
    }
    if(f==-1.0){rep(i,0,nown-1)u[i]/=nown;}
}
int main()
{
    scanf("%d%d%s%s",&l1,&l2,s1,s2);
    reverse(s1,s1+l1);
    rep(i,0,l1-1)a[i]=s1[i]=='*'?0:(s1[i]-'a'+1);
    rep(i,0,l2-1)b[i]=s2[i]=='*'?0:(s2[i]-'a'+1);
    getlen(max(l1,l2)*4);
    rep(i,0,nown-1)A[i]=a[i]*a[i]*a[i],B[i]=b[i];
    dft(A,1),dft(B,1);
    rep(i,0,nown-1)c[i]=A[i]*B[i];
    rep(i,0,nown-1)A[i]=a[i],B[i]=b[i]*b[i]*b[i];
    dft(A,1),dft(B,1);
    rep(i,0,nown-1)c[i]+=A[i]*B[i];
    rep(i,0,nown-1)A[i]=a[i]*a[i],B[i]=b[i]*b[i];
    dft(A,1),dft(B,1);
    rep(i,0,nown-1)c[i]-=2.0*A[i]*B[i];
    dft(c,-1);
    rep(i,l1-1,l2-1)if(((int)(c[i].real()+0.5))==0)ans[++num]=i-l1+2;
    write(num);puts("");
    rep(i,1,num)write(ans[i]);
    return 0;
}

转载于:https://www.cnblogs.com/xzyf/p/11598961.html