>Link

ybtoj回文子串

---

>解题思路

寻找回文子串，我们可以从这个子串的中心逐渐向外扩散，如果左边的子串等于右边的子串倒过来，就说明这个子串为回文子串。
所以我们枚举每一个点为中心，二分查找最长的扩散长度，注意这里要处理子串长度为奇数和偶数的不同情况，特殊的，如果长度为偶数，中心点是没有字符的。
如何判断“左边的子串等于右边的子串倒过来”呢？其实直接把这个字符串正过来、倒过来分别做一个哈希就行了

时间复杂度大概为 O ( n l o g n ) O(nlogn) O(nlogn)

---

>代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 1000010
#define ull unsigned long long
using namespace std;

int C, n, l, r, mid, ans;
ull a[N], hf[N], hb[N];
string s;

void find ()
{
	for (int i = 1; i <= n; i++)
	{
 		l = 1, r = n;
		while (l <= r)
		{
			mid = (l + r) / 2;
			if (i + mid > n || i - mid < 1) {r = mid - 1; continue;}
            if (hf[i - 1] - hf[i - 1 - mid] * a[mid]
			    == hb[i + 1] - hb[i + 1 + mid] * a[mid])
			{
				ans = max (ans, 2 * mid + 1);
				l = mid + 1;
			}
			else r = mid - 1;
		}
		l = 1, r = n;
		while (l <= r)
		{
			mid = (l + r) / 2;
			if (i + mid > n || i - mid + 1 < 1) {r = mid - 1; continue;}
			if (hf[i] - hf[i - mid] * a[mid]
			    == hb[i + 1] - hb[i + 1 + mid] * a[mid])
			{
				ans = max (ans, 2 * mid);
				l = mid + 1;
			}
			else r = mid - 1;
		}
	}
}

int main()
{
	cin >> s;
	a[0] = 1;
	for (int i = 1; i <= 1000000; i++) a[i] = a[i - 1] * 109;
	while (s != "END")
	{
		memset (hf, 0, sizeof (hf));
		memset (hb, 0, sizeof (hb));
		C++; ans = 1;
		n = s.size();
		s = " " + s;
		for (int i = 1; i <= n; i++)
		  hf[i] = hf[i - 1] * 109 + s[i] - 'a' + 1;
		for (int i = n; i >= 1; i--)
		  hb[i] = hb[i + 1] * 109 + s[i] - 'a' + 1;
		find ();
		printf ("Case %d: %d\n", C, ans);
		cin >> s;
	}
	return 0;
}