洛谷P5256 编程作业

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题解

这里可以学习用多项式卷积匹配的思想，每种字母分开来看

大写字母单独搞

每种小写字母，如果这个位置有这个小写字母，就是1，否则就是0

然后用哈希判断一下是否相同

字符集大小假设是$\Omega$

那么时间复杂度就是$O(\Omega (n+m))$

代码

#include 
#include 
#include 
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct Easymath
{
    ll prime[maxn], phi[maxn], mu[maxn];
    bool mark[maxn];
    ll fastpow(ll a, ll b, ll c)
    {
        ll t(a%c), ans(1ll);
        for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
        return ans;
    }
    void exgcd(ll a, ll b, ll &x, ll &y)
    {
        if(!b){x=1,y=0;return;}
        ll xx, yy;
        exgcd(b,a%b,xx,yy);
        x=yy, y=xx-a/b*yy;
    }
    ll inv(ll x, ll p)  //p是素数
    {return fastpow(x%p,p-2,p);}
    ll inv2(ll a, ll p)
    {
        ll x, y;
        exgcd(a,p,x,y);
        return (x+p)%p;
    }
    void shai(ll N)
    {
        ll i, j;
        for(i=2;i<=N;i++)mark[i]=false;
        *prime=0;
        phi[1]=mu[1]=1;
        for(i=2;i<=N;i++)
        {
            if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
            for(j=1;j<=*prime and i*prime[j]<=N;j++)
            {
                mark[i*prime[j]]=true;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
}em;
#define mod 998244353ll
#define Base 4894651ll
ll inv[maxn], PB[maxn];
struct Hash
{
    ll M, base;   //m should be a prime number, and base should be less than m.
    vector<int> h;
    Hash(ll M, ll base)
    {
        this->M=M;
        this->base=base;
    }
    void build(char* s, ll len)
    {
        ll i;
        h.resize(len+1);
        rep(i,1,len)
        {
            if(s[i])h[i]=(h[i-1]+s[i]*PB[i-1])%M;
            else h[i]=h[i-1];
        }
    }
    ll substr(ll l, ll r)   //返回非负哈希值
    {
        return (((ll)h[r]-h[l-1])*inv[l-1]%M+M)%M;
    }
};
ll n;
char s[maxn], t[maxn];
char ss[27][maxn], tt[27][maxn];
int main()
{
    ll i;
    inv[0]=1;
    inv[1]=em.inv(Base,mod);
    rep(i,2,maxn-1)inv[i]=inv[i-1]*inv[1]%mod;
    PB[0]=1;
    rep(i,1,maxn-1)PB[i]=PB[i-1]*Base%mod;
    ll T=read();
    while(T--)
    {
        scanf("%s%s",s+1,t+1);
        ll n=strlen(s+1), m=strlen(t+1), i, j, ans=0;
        
        cl(ss), cl(tt);
        rep(i,1,n)if(s[i]>'z' or s[i]<'a')ss[0][i]=s[i];
        rep(j,1,26)rep(i,1,n)ss[j][i]=(s[i]=='a'+j-1);
        rep(i,1,n)if(t[i]>'z' or t[i]<'a')tt[0][i]=t[i];
        rep(j,1,26)rep(i,1,n)tt[j][i]=(t[i]=='a'+j-1);
        
        Hash *hs[27], *ht[27];
        rep(i,0,26)
        {
            hs[i] = new Hash(998244353,4894651);
            ht[i] = new Hash(998244353,4894651);
            hs[i]->build(ss[i],n);
            ht[i]->build(tt[i],m);
        }
        ll f[27];
        memset(f,-1,sizeof(f));
        drep(i,m,1)
        {
            if('a'<=t[i] and t[i]<='z')f[t[i]-'a'+1]=i;
        }
        rep(i,1,n-m+1)
        {
            bool ok = true;
            if(hs[0]->substr(i,i+m-1)!=ht[0]->substr(1,m))ok=false;
            rep(j,1,26)
            {
                if(!ok)break;
                if(f[j]==-1)continue;
                ll c=s[i+f[j]-1]-'a'+1;
                if(hs[c]->substr(i,i+m-1)!=ht[j]->substr(1,m))ok=false;
            }
            ans+=ok;
        }
        printf("%lld\n",ans);
        rep(i,0,26)
        {
            delete hs[i];
            delete ht[i];
        }
    }
    return 0;
}