I Love Palindrome String HDU - 6599 回文树+hash

题意：

输出每个长度下的回文串(这些回文串的左半边也需要是回文串)

题解：

直接套上回文树，然后每找到一个回文串就将他hash。

因为符合要求的回文串本身是回文串，左半边也是回文串，所以它左半边也右半边相同，

自己画个图很容易发现的

每次hash判断一下就好了

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <unordered_map>

 #define  pi    acos(-1.0)
 #define  eps   1e-9
 #define  fi    first
 #define  se    second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug                printf("******\n")
 #define  mem(a, b)          memset(a,b,sizeof(a))
 #define  name2str(x)        #x
 #define  fuck(x)            cout<<#x" = "<<x<<endl
 #define  sfi(a)             scanf("%d", &a)
 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  sfL(a)             scanf("%lld", &a)
 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 #define  sfs(a)             scanf("%s", a)
 #define  sffs(a, b)         scanf("%s %s", a, b)
 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 #define  FIN                freopen("../in.txt","r",stdin)
 #define  gcd(a, b)          __gcd(a,b)
 #define  lowbit(x)          x&-x
 #define  IO                 iOS::sync_with_stdio(false)

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const ULL seed = ;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 const int maxn = 1e6 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;
 char s[maxn];
 ULL p[maxn], HA[maxn];
 int ans[maxn];

 ULL get_HASH(int l, int r) {
     return HA[r] - HA[l - ] * p[r - l + ];
 }

 struct Palindrome_Automaton {
     int len[maxn], next[maxn][], fail[maxn], cnt[maxn];
     int num[maxn], str[maxn], sz, n, last;
     int vis[maxn];

     int newnode(int l) {
         for (int i = ; i < ; ++i)next[sz][i] = ;
         cnt[sz] = num[sz] = , len[sz] = l;
         return sz++;
     }

     void init() {
         sz = n = last = ;
         newnode();
         newnode(-);
         str[] = -;
         fail[] = ;
         mem(vis, );
     }

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

     void add(int c, int pos) {
         c -= 'a';
         str[++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;
             num[now] = num[fail[now]] + ;
             int temp = (len[now] + ) / ;
             // fuck(n - len[now] + 1),fuck(n - len[now] + temp),fuck(n - temp),fuck(n);
             if (len[now] ==  ||
                 get_HASH(n - len[now] + , n - len[now] + temp) == get_HASH(n - temp+, n))
                 vis[now] = ;
             else vis[now] = ;
         }
         last = next[cur][c];
         cnt[last]++;
     }

     void count()//统计本质相同的回文串的出现次数
     {
         for (int i = sz - ; i >= ; --i) {
             cnt[fail[i]] += cnt[i];
             if (vis[i]) ans[len[i]] += cnt[i];
         }
         //逆序累加，保证每个点都会比它的父亲节点先算完，于是父亲节点能加到所有子孙
     }
 } pam;

 int main() {
 //    FIN;
     p[] = ;
     for (int i = ; i < maxn; ++i) p[i] = p[i - ] * seed;
     while (~sfs(s + )) {
         int n = strlen(s + );
         pam.init();
         for (int i = ; i <= n; ++i) {
             HA[i] = HA[i - ] * seed + s[i];
             pam.add(s[i], i);
             ans[i] = ;
         }
         pam.count();
         for (int i = ; i <= n; ++i) printf("%d%c", ans[i], (i == n ? '\n' : ' '));
     }
     return ;
 }