[2019杭电多校第二场][hdu6599]I Love Palindrome String(回文自动机&&hash)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6599

题目大意为求字符串S有多少个子串S[l,r]满足回文串的定义，并且S[l,(l+r)/2]也满足回文串的定义。

可以直接建回文自动机，然后再统计出每种回文串的个数，然后再枚举状态，判断该状态所表示的回文串它的一半是否满足回文串的定义，判断用hash来判断即可。

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<string>
 #include<algorithm>
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 const int maxn = 3e5 + ;
 ull Hash[maxn], xp[maxn];
 void init() {
     xp[] = ;
     for (int i = ; i <= 3e5 + ; i++)
         xp[i] = xp[i - ] * ;
 }
 ull getH(int l, int r) {
     if (l == )return Hash[r];
     return Hash[r] - Hash[l - ] * xp[r - l + ];
 }
 bool check(int l, int r) {
     int len = r - l + ;
     int mid = l + r >> ;
     if (len & )return getH(l, mid) == getH(mid, r);
     else return getH(l, mid) == getH(mid + , r);
 }
 ll ans[maxn], id[maxn];
 struct Palindromic_Tree {
     int next[maxn][],fail[maxn], cnt[maxn],len[maxn],S[maxn];
     int last, n,p;
     int newnode(int l) {
         for (int i = ; i < ; ++i) next[p][i] = ;
         cnt[p] = ;
         len[p] = l;
         return p++;
     }
     void init() {
         p = ;
         newnode();
         newnode(-);
         last = ;
         n = ;
         S[n] = -;
         fail[] = ;
     }

     int get_fail(int x) {
         while (S[n - len[x] - ] != S[n]) x = fail[x];
         return x;
     }
     void add(int c) {
         c -= 'a';
         S[++n] = c;
         int cur = get_fail(last);
         if (!next[cur][c]) {
             int now = newnode(len[cur] + );
             fail[now] = next[get_fail(fail[cur])][c];
             next[cur][c] = now;
         }
         last = next[cur][c];
         cnt[last]++;
         id[last] = n;
     }
     void count() {
         for (int i = p - ; i >= ; --i) cnt[fail[i]] += cnt[i];
         for (int i = ; i < p; i++)
             if (check(id[i] - len[i], id[i] - ))
                 ans[len[i]] += cnt[i];
     }
 }a;
 char s[maxn];
 int main() {
     init();
     while (~scanf("%s", s)) {
         int n = strlen(s);
         Hash[] = s[];
         memset(ans, , sizeof(ans));
         for (int i = ; i < n; i++)
             Hash[i] = Hash[i - ] *  + s[i];
         a.init();
         for (int i = ; i < n; i++)
             a.add(s[i]);
         a.count();
         for (int i = ; i <= n; i++)
             printf("%lld%c", ans[i], (i == n ? '\n' : ' '));
     }
 }