回文自动机pam
目的：类似回文Trie树+ac自动机，可以用来统计一些其他的回文串相关的量
复杂度：O(nlogn)
https://blog.csdn.net/Lolierl/article/details/99971257
https://www.luogu.org/problem/P5496
求出以每个位置结尾的回文子串个数，强制在线
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int maxn=2e6+10; struct pam_trie { int ch[26]; int fail,len,num; }; struct pam { pam_trie b[maxn]; int n,length,last,cnt,s[maxn]; char c[maxn]; pam() { b[0].len=0;b[1].len=-1; b[0].fail=1;b[1].fail=0; last=0; cnt=1; } void read() { scanf("%s",c+1); length=strlen(c+1); } int get_fail(int x) { while(s[n-b[x].len-1]!=s[n])x=b[x].fail; return x; } void insert() { int p=get_fail(last); if(!b[p].ch[s[n]]) { b[++cnt].len=b[p].len+2; b[cnt].fail=b[get_fail(b[p].fail)].ch[s[n]]; //b[cnt].num=b[b[cnt].fail].num+1; b[p].ch[s[n]]=cnt; } last=b[p].ch[s[n]]; b[last].num=b[b[last].fail].num+1; } void solve() { int k=0; s[0]=26; for(n=1;n<=length;n++) { c[n]=(c[n]-97+k)%26+97; s[n]=c[n]-'a'; insert(); printf("%d ",b[last].num); k=b[last].num; } } }P; int main() { P.read(); P.solve(); return 0; }
https://www.luogu.org/problem/P3649
求回文子串出现次数*长度的最大值
#include<iostream> #include<stdio.h> #include<cstdio> #include<cstring> using namespace std; const int maxn=3e5+10; struct pam_trie { int ch[26]; int fail,len,sum; }; struct pam { pam_trie b[maxn]; int n,length,last,cnt,s[maxn]; char c[maxn]; long long ans; pam() { b[0].len=0;b[1].len=-1; b[0].fail=1;b[1].fail=0; last=0; cnt=1; } void read() { scanf("%s",c+1); length=strlen(c+1); } int get_fail(int x) { while(s[n-b[x].len-1]!=s[n])x=b[x].fail; return x; } void insert() { int p=get_fail(last); if(!b[p].ch[s[n]]) { b[++cnt].len=b[p].len+2; b[cnt].fail=b[get_fail(b[p].fail)].ch[s[n]]; b[p].ch[s[n]]=cnt; } last=b[p].ch[s[n]]; b[last].sum++; } void solve() { s[0]=26; for(n=1;n<=length;n++) { s[n]=c[n]-'a'; insert(); } ans=0; for(int i=cnt;i>0;i--) { b[b[i].fail].sum+=b[i].sum; ans=max(ans,1ll*b[i].sum*b[i].len); } printf("%lld\n",ans); } }P; int main() { P.read(); P.solve(); }
https://www.luogu.org/problem/P4287
计算串的最长双倍回文子串的长度，tips：fail指针指向当前节点所表示的回文串的最长回文后缀
#include<stdio.h> #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int maxn=5e5+10; struct pam_trie { int ch[26]; int fail,len,sum; }; struct pam { pam_trie b[maxn]; int n,length,last,cnt,s[maxn]; char c[maxn]; pam() { b[0].len=0;b[1].len=-1; b[0].fail=1;b[1].fail=0; last=0;cnt=1; } void read() { scanf("%d",&length); scanf("%s",c+1); } int get_fail(int x) { while(s[n-b[x].len-1]!=s[n])x=b[x].fail; return x; } void insert() { int p=get_fail(last); if(!b[p].ch[s[n]]) { b[++cnt].len=b[p].len+2; b[cnt].fail=b[get_fail(b[p].fail)].ch[s[n]]; b[p].ch[s[n]]=cnt; } last=b[p].ch[s[n]]; b[last].sum++; } void solve() { s[0]=26; for(n=1;n<=length;n++) { s[n]=c[n]-'a'; insert(); } int ans=0; for(int i=cnt;i>0;i--) { int pos=i; if(b[i].len%4!=0||b[i].len<=ans)continue; while(2*b[pos].len>b[i].len)pos=b[pos].fail; if(2*b[pos].len==b[i].len)ans=b[i].len; } printf("%d\n",ans); } }P; int main() { P.read(); P.solve(); return 0; }
ICPC 2018 南京 Mediocre String Problem，用回文自动机来求出以每个位置结尾的回文子串个数，再进行exkmp
#include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int maxn=1e6+10; struct pam_trie { int ch[26]; int fail,len,num; }; int res[maxn]; char s1[maxn],s2[maxn],t[maxn]; struct pam { pam_trie b[maxn]; int n,last,cnt,s[maxn],length; pam() { b[0].len=0;b[1].len=-1; b[0].fail=1;b[1].fail=0; last=0; cnt=1; } int get_fail(int x) { while(s[n-b[x].len-1]!=s[n])x=b[x].fail; return x; } int insert() { int p=get_fail(last); if(!b[p].ch[s[n]]) { b[++cnt].len=b[p].len+2; b[cnt].fail=b[get_fail(b[p].fail)].ch[s[n]]; b[p].ch[s[n]]=cnt; } last=b[p].ch[s[n]]; b[last].num=b[b[last].fail].num+1; return b[last].num; } void solve() { s[0]=26; length=strlen(s1); for(n=1;n<=length;n++) { s[n]=s1[n-1]-'a'; res[n-1]=insert(); } } }P; int Next[maxn],extend[maxn]; void get_next(char *s) { int n=strlen(s),i,j,k=1; for(j=0;1+j<n&&s[j]==s[1+j];j++); Next[1]=j; for(i=2;i<n;i++) { int len=k+Next[k],L=Next[i-k]; if(L<len-i)Next[i]=L; else { for(j=max(0,len-i);i+j<n&&s[j]==s[i+j];j++); Next[i]=j; k=i; } } Next[0]=n; } void ex_kmp(char *T,char *s) { int n=strlen(T),m=strlen(s),i,j,k; for(j=0;j<n&&j<m&&T[j]==s[j];j++); extend[0]=j; k=0; for(i=1;i<n;i++) { int len=k+extend[k],L=Next[i-k]; if(L<len-i)extend[i]=L; else { for(j=max(0,len-i);j<m&&i+j<n&&s[j]==T[i+j];j++); extend[i]=j; k=i; } } } int main() { scanf("%s",s1); scanf("%s",t); int lens=strlen(s1); reverse(s1,s1+lens); for(int i=0;i<lens;i++)s2[i]=s1[i]; s2[lens]='\0'; get_next(t); ex_kmp(s2,t); long long ans=0; P.solve(); for(int i=1;i<lens;i++) { ans+=1ll*extend[i]*res[i-1]; } printf("%lld\n",ans); return 0; }
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