HDU 5157 Harry and magic string
Problem Description
Harry got a string T, he wanted to know the number of T’s disjoint palindrome substring pairs. A string is considered to be palindrome if and only if it reads the same backward or forward. For two substrings of
T:x=T[a1…b1],y=T[a2…b2] (where a1 is the beginning index of
x,b1 is the ending index of x.
a2,b2 as the same of y), if both x and y are palindromes and
b1<a2 or b2<a1 then we consider (x, y) to be a disjoint palindrome substring pair of T.
Input
There are several cases.
For each test case, there is a string T in the first line, which is composed by lowercase characters. The length of T is in the range of [1,100000].
For each test case, there is a string T in the first line, which is composed by lowercase characters. The length of T is in the range of [1,100000].
Output
For each test case, output one number in a line, indecates the answer.
Sample Input
aca aaaa
Sample Output
3 15HintFor the first test case there are 4 palindrome substrings of T. They are: S1=T[0,0] S2=T[0,2] S3=T[1,1] S4=T[2,2] And there are 3 disjoint palindrome substring pairs. They are: (S1,S3) (S1,S4) (S3,S4). So the answer is 3.
求有多少对回文串不想交，回文树前后两遍统计即可，用链表建边牺牲一点时间省了很多的空间。
#pragma comment(linker, "/STACK:102400000,102400000")
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int maxn = 1e5 + 10;
LL a[maxn];
char s[maxn];
struct linklist
{
int nt[maxn], ft[maxn], u[maxn], v[maxn], sz;
void clear() { sz = 0; }
void clear(int x) { ft[x] = -1; }
int get(int x, int y)
{
for (int i = ft[x]; i != -1; i = nt[i])
{
if (u[i] == y) return v[i];
}
return 0;
}
void insert(int x, int y, int z)
{
u[sz] = y; v[sz] = z;
nt[sz] = ft[x]; ft[x] = sz++;
}
};
struct PalindromicTree
{
const static int maxn = 1e5 + 10;
const static int size = 26;
linklist next;
int sz, tot, last;
int fail[maxn], len[maxn], cnt[maxn];
char s[maxn];
void clear()
{
len[1] = -1; len[2] = 0;
fail[1] = fail[2] = 1;
cnt[1] = cnt[2] = tot = 0;
last = (sz = 3) - 1;
next.clear(); next.clear(1); next.clear(2);
}
int Node(int length)
{
len[sz] = length;
cnt[sz] = 1; next.clear(sz);
return sz++;
}
int getfail(int x)
{
while (s[tot] != s[tot - len[x] - 1]) x = fail[x];
return x;
}
int add(char pos)
{
int x = (s[++tot] = pos) - 'a', y = getfail(last);
if (!(last = next.get(y, x)))
{
next.insert(y, x, last = Node(len[y] + 2));
fail[last] = len[last] == 1 ? 2 : next.get(getfail(fail[y]), x);
cnt[last] += cnt[fail[last]];
}
return cnt[last];
}
}solve;
int main()
{
while (scanf("%s", s) != EOF)
{
LL ans = a[0] = 0, len = strlen(s);
solve.clear();
for (int i = 1; i <= len; i++)
{
a[i] = a[i - 1] + solve.add(s[i - 1]);
}
solve.clear();
for (int i = len; i; i--)
{
ans += solve.add(s[i - 1])*a[i - 1];
}
printf("%lld\n", ans);
}
return 0;
}