Victor and String HDU - 5421 双向回文树

题意：

有n种操作，开始给你一个空串，给你4中操作

1 c  在字符串的首部添加字符c

2 c  在字符串的尾部添加字符c

3  询问字符中的本质不同的回文串的个数

4 询问字符串中回文串的个数

思路：last[0]表示首部的操作的位置，last[1]表示尾部的操作的位置

模板提，用上双向的回文树就好了。

 #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 = 1e5 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;
 char s[maxn];
 LL sum[maxn];

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

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

     void init() {
         sz = ;
         newnode();
         newnode(-);
         S[] = -;
         fail[] = ;
         last[] = last[] = ;
         n[] = maxn - , n[] = maxn - ;
     }

     int get_fail(int x, int k) {
         S[n[] - ] = -, S[n[] + ] = -;
         while (S[n[k] - (k ?  : -) * (len[x] + )] != S[n[k]])x = fail[x];
         return x;
     }

     int add(int c, int k) {
         c -= 'a';
         S[n[k] += (k ?  : -)] = c;
         int cur = get_fail(last[k],k);
         if (!(last[k] = next[cur][c])) {
             int now = newnode(len[cur] + );
             fail[now] = next[get_fail(fail[cur],k)][c];
             next[cur][c] = now;
             num[now] = num[fail[now]] + ;
             last[k]=now;
             if (len[last[k]]==n[]-n[]+) last[k^]=last[k];
         }
         //last[k] = next[cur][c];
         //cnt[last]++;
         return num[last[k]];
     }

     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];
         //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;
 char op[];
 int n;
 int main() {
     //FIN;
     while(~sfi(n)){
         pam.init();
         LL ans=;
         for (int i = ,x; i < n; ++i) {
             sfi(x);
             if (x==) {
                 sfs(op);
                 ans+=pam.add(op[],);
             }else if (x==) {
                 sfs(op);
                 ans+=pam.add(op[],);
             }else if (x==) printf("%d\n",pam.sz-);
             else printf("%lld\n",ans);
         }
     }
     return ;
 }