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
1 #include <set>
2 #include <map>
3 #include <stack>
4 #include <queue>
5 #include <cmath>
6 #include <ctime>
7 #include <cstdio>
8 #include <string>
9 #include <vector>
10 #include <cstring>
11 #include <iostream>
12 #include <algorithm>
13 #include <unordered_map>
14
15 #define pi acos(-1.0)
16 #define eps 1e-9
17 #define fi first
18 #define se second
19 #define rtl rt<<1
20 #define rtr rt<<1|1
21 #define bug printf("******\n")
22 #define mem(a, b) memset(a,b,sizeof(a))
23 #define name2str(x) #x
24 #define fuck(x) cout<<#x" = "<<x<<endl
25 #define sfi(a) scanf("%d", &a)
26 #define sffi(a, b) scanf("%d %d", &a, &b)
27 #define sfffi(a, b, c) scanf("%d %d %d", &a, &b, &c)
28 #define sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
29 #define sfL(a) scanf("%lld", &a)
30 #define sffL(a, b) scanf("%lld %lld", &a, &b)
31 #define sfffL(a, b, c) scanf("%lld %lld %lld", &a, &b, &c)
32 #define sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
33 #define sfs(a) scanf("%s", a)
34 #define sffs(a, b) scanf("%s %s", a, b)
35 #define sfffs(a, b, c) scanf("%s %s %s", a, b, c)
36 #define sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
37 #define FIN freopen("../in.txt","r",stdin)
38 #define gcd(a, b) __gcd(a,b)
39 #define lowbit(x) x&-x
40 #define IO iOS::sync_with_stdio(false)
41
42
43 using namespace std;
44 typedef long long LL;
45 typedef unsigned long long ULL;
46 const ULL seed = 13331;
47 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
48 const int maxn = 1e6 + 7;
49 const int maxm = 8e6 + 10;
50 const int INF = 0x3f3f3f3f;
51 const int mod = 1e9 + 7;
52 char s[maxn];
53 ULL p[maxn], HA[maxn];
54 int ans[maxn];
55
56 ULL get_HASH(int l, int r) {
57 return HA[r] - HA[l - 1] * p[r - l + 1];
58 }
59
60 struct Palindrome_Automaton {
61 int len[maxn], next[maxn][26], fail[maxn], cnt[maxn];
62 int num[maxn], str[maxn], sz, n, last;
63 int vis[maxn];
64
65 int newnode(int l) {
66 for (int i = 0; i < 26; ++i)next[sz][i] = 0;
67 cnt[sz] = num[sz] = 0, len[sz] = l;
68 return sz++;
69 }
70
71 void init() {
72 sz = n = last = 0;
73 newnode(0);
74 newnode(-1);
75 str[0] = -1;
76 fail[0] = 1;
77 mem(vis, 0);
78 }
79
80 int get_fail(int x) {
81 while (str[n - len[x] - 1] != str[n])x = fail[x];
82 return x;
83 }
84
85 void add(int c, int pos) {
86 c -= 'a';
87 str[++n] = c;
88 int cur = get_fail(last);
89 if (!next[cur][c]) {
90 int now = newnode(len[cur] + 2);
91 fail[now] = next[get_fail(fail[cur])][c];
92 next[cur][c] = now;
93 num[now] = num[fail[now]] + 1;
94 int temp = (len[now] + 1) / 2;
95 // fuck(n - len[now] + 1),fuck(n - len[now] + temp),fuck(n - temp),fuck(n);
96 if (len[now] == 1 ||
97 get_HASH(n - len[now] + 1, n - len[now] + temp) == get_HASH(n - temp+1, n))
98 vis[now] = 1;
99 else vis[now] = 0;
100 }
101 last = next[cur][c];
102 cnt[last]++;
103 }
104
105 void count()//统计本质相同的回文串的出现次数
106 {
107 for (int i = sz - 1; i >= 0; --i) {
108 cnt[fail[i]] += cnt[i];
109 if (vis[i]) ans[len[i]] += cnt[i];
110 }
111 //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
112 }
113 } pam;
114
115 int main() {
116 // FIN;
117 p[0] = 1;
118 for (int i = 1; i < maxn; ++i) p[i] = p[i - 1] * seed;
119 while (~sfs(s + 1)) {
120 int n = strlen(s + 1);
121 pam.init();
122 for (int i = 1; i <= n; ++i) {
123 HA[i] = HA[i - 1] * seed + s[i];
124 pam.add(s[i], i);
125 ans[i] = 0;
126 }
127 pam.count();
128 for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], (i == n ? '\n' : ' '));
129 }
130 return 0;
131 }
View Code