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蒟蒻懒得学马拉车,于是就用hash写了。

预处理出每个点作为回文串的左端点和右端点时回文串最长为多少

然后枚举断点,统计答案就可以了

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;i++)
    #define Fdr(i,x,y) for(int i=x;i>=y;i--)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) (int(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 100011
char s[N];
ull h1[N],h2[N],p[N];
int n,ans=0;
int L[N],R[N];
ull gh1(int l,int r){
    return h1[r]-h1[l-1]*p[r-l+1];
}
ull gh2(int l,int r){
    return h2[l]-h2[r+1]*p[r-l+1];
}
void a1(int x){
    int l=1,r=MIN(x,n-x);
    while(l<=r){
        int m=(l+r)>>1;
        if(gh1(x-m,x+m)==gh2(x-m,x+m))
            l=m+1;
        else r=m-1;
    }
    L[x+r]=MAX(L[x+r],r*2+1);
    R[x-r]=MAX(R[x-r],r*2+1);
}
void a2(int x){
    int l=1,r=MIN(x,n-x);
    while(l<=r){
        int m=(l+r)>>1;
        if(gh1(x-m+1,x+m)==gh2(x-m+1,x+m))
            l=m+1;
        else r=m-1;
    }
    L[x+r]=MAX(L[x+r],r*2);
    R[x-r+1]=MAX(R[x-r+1],r*2);
}
#define base 233
int main(){
    in>>(s+1);
    p[0]=1;
    n=strlen(s+1);
    Fur(i,1,n){
        L[i]=R[i]=1;
        h1[i]=h1[i-1]*base+s[i];
        p[i]=p[i-1]*base;
    }
    Fdr(i,n,1)
        h2[i]=h2[i+1]*base+s[i];

    Fur(i,1,n)a1(i),a2(i);

    Fur(i,1,n)R[i]=MAX(R[i],R[i-1]-2);
    Fdr(i,n,1)L[i]=MAX(L[i],L[i+1]-2);

    Fur(i,1,n-1)ans=MAX(ans,L[i]+R[i+1]);
    out<<ans<<ln;
}
LG 4555 [国家集训队]最长双回文串
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