The question

Determine whether the string can be S It is divided into three non empty palindromes


manacher Preprocess the position of prefix and suffix palindrome , Enumerates the first palindrome string and the third palindrome string , So we get the interval of the second palindrome string , Find the midpoint , because manacher After processing, all palindrome strings are odd in length , Then judge whether the middle part is palindrome or not according to the palindrome radius of the midpoint , Complexity o(n2). as for n2 Why can complexity pass .. Is not very good

#define rep(i,e) for(int i=0;i<(e);i++)
#define rep1(i,e) for(int i=1;i<=(e);i++)
#define repx(i,x,e) for(int i=(x);i<=(e);i++)
#define X first
#define Y second
#define PB push_back
#define MP make_pair
#define mset(var,val) memset(var,val,sizeof(var))
#define scd(a) scanf("%d",&a)
#define scdd(a,b) scanf("%d%d",&a,&b)
#define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define pd(a) printf("%d\n",a)
#define scl(a) scanf("%lld",&a)
#define scll(a,b) scanf("%lld%lld",&a,&b)
#define sclll(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)
#define IOS ios::sync_with_stdio(false);cin.tie(0) using namespace std;
typedef long long ll;
template <class T>
void test(T a){cout<<a<<endl;}
template <class T,class T2>
void test(T a,T2 b){cout<<a<<" "<<b<<endl;}
template <class T,class T2,class T3>
void test(T a,T2 b,T3 c){cout<<a<<" "<<b<<" "<<c<<endl;}
const int N = 1e6+;
//const int MAXN = 210;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const ll mod = ;
int T;
void testcase(){
printf("Case #%d: ",++T);
const int MAXN = 4e4+;
const int MAXM = ;
char ma[MAXN],s[MAXN];
int mp[MAXN];
int pre[MAXN],suf[MAXN];
void Manacher(int len){
int l=;
for(int i=;i<len;i++) ma[l++]=s[i],ma[l++]='#';
int mx=,id=;
for(int i=;i<l;i++){
while(ma[i+mp[i]]==ma[i-mp[i]]) mp[i]++;
int main() {
#ifdef LOCAL
#endif // LOCAL
int t;
int len = strlen(s);
int l=,r=;
for(int i=;i<=len;i++){
if(mp[i]==i&&i!=) pre[l++]=mp[i];
//mp[i]==i Guarantee mp[i] Can be used as the radius of the first palindrome string , Join in pre Array ,i!=1 Make sure that the first palindrome string is not empty
if(mp[i]+i-==len&&i!=len) suf[r++]=mp[i];
int i,j;
int t1=*pre[i];
int t2=len+-*suf[j];
int tmp = (t1+t2)>>;
if(t1>t2) continue;
if(mp[tmp]==) continue;
if(mp[tmp]*->=t2-t1+) break;
if(j<r) break;
if(i>=) puts("Yes");
else puts("No");
return ;

HDU - 5340 Three Palindromes(manacher Algorithm ) More articles about

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