BZOJ2565【回文自动机】

跑两遍PAM.分别找到每个点做为结尾字符和开始字符的最长回文串的长度.

/* I will wait for you */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;

const int maxn=500010;
const int maxm=1010;
const int maxs=26;
const int INF=1<<29;
const int P=1000000007;
const double error=1e-9;

inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9') f=(ch=='-'?-1:1),ch=getchar();
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x*f;
}

struct pam
{
pam *fa,*next[maxs];int len,val;
}su[maxn],*headf,*heads,*last;

int n,pos,cnt,ans,l[maxn],r[maxn];char s[maxn];

void init()
{
s[0]='~',pos=0;
headf=&su[++cnt],last=heads=&su[++cnt];
headf->fa=heads,heads->len=-1;
}

void add(int x)
{
pam *per=last;
for(pos++;s[pos-per->len-1]!=s[pos];per=per->fa);

if(!per->next[x])
{
pam *now=&su[++cnt];
last=per->next[x]=now,now->len=per->len+2;
if(per==heads) now->fa=headf;
else
{
for(per=per->fa;s[pos-per->len-1]!=s[pos];per=per->fa);
now->fa=per->next[x];
}
}
else last=per->next[x];
}

int main()
{
scanf("%s",s+1),n=strlen(s+1);

init();
for(int i=1;i<=n;i++) add(s[i]-'a'),l[i]=last->len;
init();
for(int i=1;i<=n/2;i++) swap(s[i],s[n-i+1]);
for(int i=1;i<=n;i++) add(s[i]-'a'),r[n-i+1]=last->len;

for(int i=1;i<n;i++) ans=max(ans,l[i]+r[i+1]);
printf("%d\n",ans);

return 0;
}

膜拜一下lowsfish的BZOJ rank 1的做法.

每次求出以当前字符String[now]做结尾字符的最长回文串的长度lenth[now].用lenth[now]+lenth[now-lenth[now]]更新答案.

/* I will wait for you */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;

const int maxn=100010;
const int maxm=1010;
const int maxs=26;
const int INF=1<<29;
const int P=1000000007;
const double error=1e-9;

inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9') f=(ch=='-'?-1:1),ch=getchar();
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x*f;
}

struct pam
{
pam *fa,*next[maxs];int len;
}su[maxn],*headf,*heads,*last;

int n,pos,cnt,ans,f[maxn];char s[maxn];

void init()
{
s[0]='~',pos=0;
headf=&su[++cnt],last=heads=&su[++cnt];
headf->fa=heads,heads->len=-1;
}

inline void add(int x)
{
pam *per=last;
for(pos++;s[pos-per->len-1]!=s[pos];per=per->fa);

if(!per->next[x])
{
pam *now=&su[++cnt];
last=per->next[x]=now,now->len=per->len+2;
if(per==heads) now->fa=headf;
else
{
for(per=per->fa;s[pos-per->len-1]!=s[pos];per=per->fa);
now->fa=per->next[x];
}
}
else last=per->next[x];
}

int main()
{
scanf("%s",s+1),n=strlen(s+1),init();

for(int i=1;i<=n;i++) add(s[i]-'a'),ans=max(ans,(f[i]=last->len)+f[i-f[i]]);

printf("%d\n",ans);
return 0;
}