2019牛客全国多校训练四 I题 string (SAM+PAM)

链接：https://ac.nowcoder.com/acm/contest/884/I
来源：牛客网

题目描述

We call a,ba,ba,b non-equivalent if and only if a≠ba \neq ba​=b and a≠rev(b)a \neq rev(b)a​=rev(b), where rev(s)rev(s)rev(s) refers to the string obtained by reversing characters of sss, for example rev(abca)=acbarev(abca)=acbarev(abca)=acba.

There is a string sss consisted of lower-case letters. You need to find some substrings of sss so that any two of them are non-equivalent. Find out what's the largest number of substrings you can choose.

输入描述:

A line containing a string sss of lower-case letters.

输出描述:

A positive integer - the largest possible number of substrings of sss that are non-equivalent.

示例1

输入

abac

输出

8

说明

The set of following substrings is such a choice: abac,b,a,ab,aba,bac,ac,cabac,b,a,ab,aba,bac,ac,cabac,b,a,ab,aba,bac,ac,c.

备注:

1≤∣s∣≤2×1051 \leq |s|\leq 2 \times 10^51≤∣s∣≤2×105, sss is consisted of lower-case letters.

题解:

题目给你一个字符串s,让你求s中的子串组成的最大集合，满足这个集合内的每一个子串str,  str和rev(str)不同时存在{rev(str):表示str反过来}

思路：就是先用SAM统计出s#rev(s)中不包含 '#'的所有子串ans1; 然后用PAM统计出s中本质不同的子串数量ans2;

这答案就是(ans1+ans2)/2;

为什么呢？

因为在用SAM统计s#rev(s)的时候会把所有字符串统计两边，而本身就是回文串的只会统计一遍。

参考代码：

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 #define pii pair<int,int>
 #define pil pair<int,long long>
 const int INF=0x3f3f3f3f;
 const ll inf=0x3f3f3f3f3f3f3f3fll;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 const int maxn=4e5+;
 const int MAXN=4e5+;
 char str[maxn];
 int s[maxn];
 ll ans;
 struct SAM{
     int l[maxn<<],fa[maxn<<],nxt[maxn<<][];
     int last,cnt;

     void Init()
     {
         ans=;last=cnt=;
         l[cnt]=fa[cnt]=;
         memset(nxt[cnt],,sizeof(nxt[cnt]));
     }

     int NewNode()
     {
         ++cnt;
         memset(nxt[cnt],,sizeof(nxt[cnt]));
         l[cnt]=fa[cnt]=;
         return cnt;
     }

     void Insert(int ch)
     {
         int np=NewNode(),p=last;
         last=np; l[np]=l[p]+;
         while(p&&!nxt[p][ch]) nxt[p][ch]=np,p=fa[p];
         if(!p) fa[np]=;
         else
         {
             int q=nxt[p][ch];
             if(l[p]+==l[q]) fa[np]=q;
             else
             {
                 int nq=NewNode();
                 memcpy(nxt[nq],nxt[q],sizeof(nxt[q]));
                 fa[nq]=fa[q];
                 l[nq]=l[p]+;
                 fa[np]=fa[q]=nq;
                 while(nxt[p][ch]==q) nxt[p][ch]=nq,p=fa[p];
             }
         }
         ans+=1ll*(l[last]-l[fa[last]]);
     }

 }sam;

 struct Palindromic_Tree{
     int next[MAXN][];
     int fail[MAXN];
     int cnt[MAXN];
     int num[MAXN];
     int len[MAXN];
     int S[MAXN];
     int last;
     int n;
     int p;

     int newnode(int l)
     {
         for(int i=;i<;++i) next[p][i]=;
         cnt[p]=;
         num[p]=;
         len[p]=l;
         return p++;
     }

     void Init()
     {
         p=;
         newnode( );
         newnode(-);
         last=;
         n=;
         S[n]=-;
         fail[]=;
     }

     int get_fail(int x)
     {
         while(S[n-len[x]-]!=S[n])x=fail[x] ;
         return x ;
     }

     void add(int c)
     {
         S[++ n]=c;
         int cur=get_fail(last) ;
         if(!next[cur][c])
         {
             int now=newnode(len[cur]+) ;
             fail[now]=next[get_fail(fail[cur])][c] ;
             next[cur][c]=now ;
             num[now]=num[fail[now]]+;
         }
         last=next[cur][c];
         cnt[last]++;
     }

     ll count()
     {
         ll res=p*1ll;
         for(int i=p-;i>=;--i) cnt[fail[i]]+=cnt[i];
         //for(int i=1;i<=p;++i) res+=cnt[i];
         //cout<<"res "<<res<<endl;
         return (res-);
     }
 } pam;

 int main()
 {
     scanf("%s",str);
     int len=strlen(str);

     sam.Init();
     for(int i=;i<len;++i) sam.Insert(str[i]-'a');
     sam.Insert();
     for(int i=len-;i>=;--i) sam.Insert(str[i]-'a');
     ans-=1ll*(len+)*(len+);
     //cout<<"ans "<<ans<<endl;
     pam.Init();
     for(int i=;i<len;++i) pam.add(str[i]-'a');
     ans=ans+pam.count();

     printf("%lld\n",(ans/2ll));

     return ;
 }