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
eq ba�=b and a≠rev(b)a
eq 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 imes 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：
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 #define pii pair<int,int> 5 #define pil pair<int,long long> 6 const int INF=0x3f3f3f3f; 7 const ll inf=0x3f3f3f3f3f3f3f3fll; 8 inline int read() 9 { 10 int x=0,f=1;char ch=getchar(); 11 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 12 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 13 return x*f; 14 } 15 const int maxn=4e5+10; 16 const int MAXN=4e5+10; 17 char str[maxn]; 18 int s[maxn]; 19 ll ans; 20 struct SAM{ 21 int l[maxn<<1],fa[maxn<<1],nxt[maxn<<1][30]; 22 int last,cnt; 23 24 void Init() 25 { 26 ans=0;last=cnt=1; 27 l[cnt]=fa[cnt]=0; 28 memset(nxt[cnt],0,sizeof(nxt[cnt])); 29 } 30 31 int NewNode() 32 { 33 ++cnt; 34 memset(nxt[cnt],0,sizeof(nxt[cnt])); 35 l[cnt]=fa[cnt]=0; 36 return cnt; 37 } 38 39 void Insert(int ch) 40 { 41 int np=NewNode(),p=last; 42 last=np; l[np]=l[p]+1; 43 while(p&&!nxt[p][ch]) nxt[p][ch]=np,p=fa[p]; 44 if(!p) fa[np]=1; 45 else 46 { 47 int q=nxt[p][ch]; 48 if(l[p]+1==l[q]) fa[np]=q; 49 else 50 { 51 int nq=NewNode(); 52 memcpy(nxt[nq],nxt[q],sizeof(nxt[q])); 53 fa[nq]=fa[q]; 54 l[nq]=l[p]+1; 55 fa[np]=fa[q]=nq; 56 while(nxt[p][ch]==q) nxt[p][ch]=nq,p=fa[p]; 57 } 58 } 59 ans+=1ll*(l[last]-l[fa[last]]); 60 } 61 62 }sam; 63 64 struct Palindromic_Tree{ 65 int next[MAXN][26]; 66 int fail[MAXN]; 67 int cnt[MAXN]; 68 int num[MAXN]; 69 int len[MAXN]; 70 int S[MAXN]; 71 int last; 72 int n; 73 int p; 74 75 int newnode(int l) 76 { 77 for(int i=0;i<26;++i) next[p][i]=0; 78 cnt[p]=0; 79 num[p]=0; 80 len[p]=l; 81 return p++; 82 } 83 84 void Init() 85 { 86 p=0; 87 newnode( 0); 88 newnode(-1); 89 last=0; 90 n=0; 91 S[n]=-1; 92 fail[0]=1; 93 } 94 95 int get_fail(int x) 96 { 97 while(S[n-len[x]-1]!=S[n])x=fail[x] ; 98 return x ; 99 } 100 101 void add(int c) 102 { 103 S[++ n]=c; 104 int cur=get_fail(last) ; 105 if(!next[cur][c]) 106 { 107 int now=newnode(len[cur]+2) ; 108 fail[now]=next[get_fail(fail[cur])][c] ; 109 next[cur][c]=now ; 110 num[now]=num[fail[now]]+1; 111 } 112 last=next[cur][c]; 113 cnt[last]++; 114 } 115 116 ll count() 117 { 118 ll res=p*1ll; 119 for(int i=p-1;i>=0;--i) cnt[fail[i]]+=cnt[i]; 120 //for(int i=1;i<=p;++i) res+=cnt[i]; 121 //cout<<"res "<<res<<endl; 122 return (res-2); 123 } 124 } pam; 125 126 int main() 127 { 128 scanf("%s",str); 129 int len=strlen(str); 130 131 sam.Init(); 132 for(int i=0;i<len;++i) sam.Insert(str[i]-'a'); 133 sam.Insert(28); 134 for(int i=len-1;i>=0;--i) sam.Insert(str[i]-'a'); 135 ans-=1ll*(len+1)*(len+1); 136 //cout<<"ans "<<ans<<endl; 137 pam.Init(); 138 for(int i=0;i<len;++i) pam.add(str[i]-'a'); 139 ans=ans+pam.count(); 140 141 printf("%lld ",(ans/2ll)); 142 143 144 return 0; 145 }