回文自动机模板题
一篇比较详细的讲解：http://blog.csdn.net/u013368721/article/details/42100363
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cstdio> 6 #include<string> 7 #include<cmath> 8 #include<ctime> 9 #include<queue> 10 #include<stack> 11 #include<map> 12 #include<set> 13 #define rre(i,r,l) for(int i=(r);i>=(l);i--) 14 #define re(i,l,r) for(int i=(l);i<=(r);i++) 15 #define Clear(a,b) memset(a,b,sizeof(a)) 16 #define inout(x) printf("%d",(x)) 17 #define douin(x) scanf("%lf",&x) 18 #define strin(x) scanf("%s",(x)) 19 #define LLin(x) scanf("%lld",&x) 20 #define op operator 21 #define CSC main 22 typedef unsigned long long ULL; 23 typedef const int cint; 24 typedef long long LL; 25 using namespace std; 26 void inin(int &ret) 27 { 28 ret=0;int f=0;char ch=getchar(); 29 while(ch<'0'||ch>'9'){if(ch=='-')f=1;ch=getchar();} 30 while(ch>='0'&&ch<='9')ret*=10,ret+=ch-'0',ch=getchar(); 31 ret=f?-ret:ret; 32 } 33 const int xxxx=600010; 34 LL ans; 35 namespace pam 36 { 37 int ch[xxxx][26],pre[xxxx],sum[xxxx],len[xxxx],ss[xxxx],num[xxxx]; 38 int last,n,ed=-1; 39 int newnode(int x) 40 { 41 len[++ed]=x; 42 return ed; 43 } 44 void init() 45 { 46 newnode(0),newnode(-1); 47 last=n=0;ss[n]=-1,pre[0]=pre[1]=1; 48 } 49 int getfail(int x) 50 { 51 while(ss[n-len[x]-1]!=ss[n])x=pre[x]; 52 return x; 53 } 54 void add(int c) 55 { 56 c-='a'; 57 ss[++n]=c; 58 int temp=getfail(last); 59 if(!ch[temp][c]) 60 { 61 int now=newnode(len[temp]+2); 62 int k=getfail(pre[temp]); 63 pre[now]=ch[k][c]; 64 ch[temp][c]=now; 65 num[now]=num[pre[now]]+1; 66 } 67 last=ch[temp][c]; 68 sum[last]++; 69 } 70 void add(char *s) 71 { 72 int limit=strlen(s); 73 re(i,0,limit-1)add(s[i]); 74 } 75 void count() 76 { 77 rre(i,ed,1)sum[pre[i]]+=sum[i]; 78 } 79 void solve() 80 { 81 rre(i,ed,1)ans=max(ans,(LL)sum[i]*len[i]); 82 printf("%lld",ans); 83 } 84 } 85 char s[xxxx]; 86 int main() 87 { 88 strin(s); 89 pam::init(); 90 pam::add(s); 91 pam::count(); 92 pam::solve(); 93 return 0; 94 }