【BZOJ3676】【APIO2014】回文串

【题目链接】

• 点击打开链接
  

【思路要点】

• 回文树模板题。
• 时间复杂度\(O(|S|)\)。

【代码】

#include
using namespace std;
#define MAXN	300005
struct Palindromic_Tree {
	int child[MAXN][26], father[MAXN];
	int depth[MAXN], cnt[MAXN];
	int size, last, len;
	char s[MAXN];
	int new_node(int length) {
		memset(child[size], 0, sizeof(child[size]));
		father[size] = 0;
		depth[size] = length;
		cnt[size] = 0;
		return size++;
	}
	void Extend(int ch, int pos) {
		int p = last;
		while (s[pos - depth[p] - 1] != s[pos])
			p = father[p];
		if (!child[p][ch]) {
			int now = new_node(depth[p] + 2), fa = father[p];
			while (s[pos - depth[fa] - 1] != s[pos])
				fa = father[fa];
			father[now] = child[fa][ch];
			if (father[now] == 0) father[now] = 1;
			child[p][ch] = now;
		}
		last = child[p][ch];
		cnt[last]++;
	}
	void init(char *x) {
		len = strlen(x + 1);
		for (int i = 1; i <= len; i++)
			s[i] = x[i];
		size = 0;
		new_node(-1);
		new_node(0);
		father[0] = father[1] = 0;
		depth[0] = -1;
		last = 1;
		for (int i = 1; i <= len; i++)
			Extend(s[i] - 'a', i);
	}
	long long getans() {
		long long ans = 0;
		for (int i = size - 1; i > 0; i--) {
			cnt[father[i]] += cnt[i];
			ans = max(ans, (long long) cnt[i] * depth[i]);
		}
		return ans;
	}
} PT;
char s[MAXN];
int main() {
	scanf("%s", s + 1);
	PT.init(s);
	printf("%lld\n", PT.getans());
	return 0;
}