codeforces17E Palisection

求母串中互相交的回文串对数。

回文树练习题。

求相交的不好求,但是我们很容易求出不相交的(精华所在!要学会逆向思维!)

用回文树求出在此处开头和结尾的回文串非别为多少,直接搞就可以了。

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

using namespace std;
typedef long long LL;
const int Mod=51123987;
const int Maxn=2000005;
char S[Maxn];
int fail[Maxn],num[Maxn],s1[Maxn],s2[Maxn];
int a[Maxn],K[Maxn],next[Maxn],len[Maxn];
int n,c,t,p,last,i,st,ans;

int fd(int x,int c){
  for (int i=a[x];i;i=next[i])
    if (K[i]==c) return i;
  return 0;
}

int newnode(int l){
  len[st] = l;
  return st++;
}

void init(){
  st = 0;
  S[0] = '#'; 
  newnode(0);
  newnode(-1);
  fail[0] = 1;
  fail[1] = 0;
  last = 0;
}

int get_fail(int x,int n){
  while (S[n-len[x]-1]!=S[n]) x=fail[x];
  return x;
}

void work(int s[]){
  memset(a,0,sizeof(a));
  init();
  for (i=1;i<=n;i++){
  	c = S[i]-'a';
  	t = get_fail(last,i);
  	if (!fd(t,c)){
  	  p = newnode(len[t]+2);
  	  fail[p] = fd( get_fail( fail[t], i ), c );
  	  next[p] = a[t]; a[t] = p; K[p] = c;
  	  num[p] = num[fail[p]]+1;
  	}
  	last = fd(t,c);
  	s[i] = num[last];
  }
}

int main(){
  freopen("17E.in","r",stdin);
  freopen("17E.out","w",stdout);
  scanf("%d\n",&n);
  scanf("%s",S+1);
  work(s1);
  for (i=1;i<=n/2;i++)
    swap(S[i],S[n-i+1]);
  work(s2);

  LL N = 0, NN;
  for (i=1;i<=n;i++)
    N = N+(LL)s1[i];
  NN = N-1;
  if (N&1) NN>>=1; else N>>=1;
  
  for (i=1;i<=n/2;i++)
    swap(s2[i],s2[n-i+1]);
  for (i=n;i>0;i--)
    s2[i] = (s2[i]+s2[i+1])%Mod;
  for (i=1;i<n;i++)
    ans = (ans+(LL)s1[i]*s2[i+1]%Mod)%Mod;

  ans = ( (LL)(NN%Mod)*(N%Mod)%Mod-ans+Mod )%Mod;
  printf("%d\n",ans);
  return 0;
}


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