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].

 

Output

For each test case, output one number in a line, indecates the answer.

 

Sample Input

acaaaaa

 

Sample Output

315
Hint
For 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;}