题目链接
题意
定义两个字符串$a, b$等价$a=b$或$a \neq rev(b)$。现有字符串$s$，问从其中所有字串选出一个集合，集合中所有串都不相交。
询问集合最大大小。
思路
贪心的想，集合最大，就是$a$的本质不同的串和$rev(a)$的本质不同的串，的和的一半，因为每一个串在$a$中对应于一个$rev(a)$中的串，但在某些情况下这两个串是一样的。没错，就是当选择的这个子串是回文串。
我们建立新串 # ，计算其中本质不同串的个数，减去那些含有’#’的串，这个结果记为$p$。
$a$中的回文串数量为$q$，那么答案就是。1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
using namespace std;
typedef long long ll;
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
const int N = 4e5 + 5;
int t1[N], t2[N], c[N];
bool cmp(int *r, int a, int b, int l)
{
return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int str[], int sa[], int rank[], int height[], int n, int m)
{
n++;
int *x = t1, *y = t2, p;
for (int i = 0; i < m; i++)
c[i] = 0;
for (int i = 0; i < n; i++)
c[x[i] = str[i]]++;
for (int i = 1; i < m; i++)
c[i] += c[i - 1];
for (int i = n - 1; i >= 0; i--)
sa[--c[x[i]]] = i;
for (int j = 1; j <= n; j <<= 1)
{
p = 0;
for (int i = n - j; i < n; i++)
{
y[p++] = i;
}
for (int i = 0; i < n; i++)
if (sa[i] >= j)
y[p++] = sa[i] - j;
for (int i = 0; i < m; i++)
c[i] = 0;
for (int i = 0; i < n; i++)
c[x[y[i]]]++;
for (int i = 1; i < m; i++)
c[i] += c[i - 1];
for (int i = n - 1; i >= 0; i--)
sa[--c[x[y[i]]]] = y[i];
swap(x, y);
p = 1;
x[sa[0]] = 0;
for (int i = 1; i < n; i++)
x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
if (p >= n)
break;
m = p;
}
int k = 0;
n--;
for (int i = 0; i <= n; i++)
rank[sa[i]] = i;
for (int i = 0; i < n; i++)
{
if (k)
k--;
int j = sa[rank[i] - 1];
while (str[i + k] == str[j + k])
k++;
height[rank[i]] = k;
}
}
int Rank[N], height[N], sa[N];
int r[N];
const int maxn = 2e5 + 5;
char a[maxn], b[maxn];
struct Palindrome_Tree
{
int n, sz, last;
int ch[maxn][27], fail[maxn], len[maxn], s[maxn];
ll cnt[maxn];
int new_node(int x)
{
for (int i = 0; i < 26; i++)
ch[sz][i] = 0;
cnt[sz] = 0;
len[sz] = x;
return sz++;
}
void init()
{
sz = 0;
new_node(0), new_node(-1);
last = 0;
n = 0;
s[0] = -1, fail[0] = 1;
fail[1] = 0, s[1] = 0;
}
int get_fail(int u)
{
while (s[n - len[u] - 1] != s[n])
u = fail[u];
return u;
}
void add(int c)
{
c -= 'a';
s[++n] = c;
int u = get_fail(last);
if (!ch[u][c])
{
int np = new_node(len[u] + 2);
fail[np] = ch[get_fail(fail[u])][c];
ch[u][c] = np;
}
last = ch[u][c];
// cnt[last]++;
}
void counnt()
{
for (int i = sz - 1; i > 1; i--)
cnt[fail[i]] += cnt[i];
}
void dfs(int u)
{
for (int i = 0; i < 26; i++)
if (ch[u][i])
{
printf("%d %d\n", u, i);
dfs(ch[u][i]);
}
}
};
Palindrome_Tree pt;
int main()
{
scanf("%s", a);
// scanf("%s", b);
int len = 0, lena = strlen(a), lenb = lena;
for (int i = 0; i < lena; i++)
b[i] = a[lena - 1 - i];
b[lena] = 0;
for (int i = 0; i < lena; i++)
r[len++] = a[i];
r[len++] = 'z' + 1;
for (int i = 0; i < lenb; i++)
r[len++] = b[i];
r[len] = 0;
da(r, sa, Rank, height, len, 128);
ll p = (len + 1) * 1ll * len / 2;
//dbg(0, sa[0], height[0], p, len);
//dbg(1, sa[1], height[1]);
for (int i = 2; i <= len; i++)
{
p -= height[i];
//dbg(i, sa[i], height[i], p);
}
p -= (lena + 1ll) * (lenb + 1ll);
//dbg(p, len);
pt.init();
for (int i = 0; i < lena; i++)
pt.add(a[i]);
//pt.counnt();
//ll q = pt.cnt[0] + pt.cnt[1];
printf("%lld\n", (p + pt.sz - 2) / 2);
return 0;
}