POI2006PAL-Palindromes

POI #Year2006 #字符串 #妙妙题 #hash

结论
对于两个回文串，当且仅当两个串的最小循环节相同，这两个串拼起来是一个回文串

那么就可以用 \(kmp\) 维护，最短循环节的长度为

• 当 \(m\mod nxt_m=0\) 时为 \(m-nxt_m\)
• 否则为 \(m\)

然后 \(hash\) 求相同的个数

// Author: xiaruize
using namespace std;
#ifndef ONLINE_JUDGE
#define debug(...) std::cerr << __LINE__ << ": [", __DEBUG_UTIL__::printer(#__VA_ARGS__, __VA_ARGS__)
#define debugArr(...) std::cerr << __LINE__ << ": [", __DEBUG_UTIL__::printerArr(#__VA_ARGS__, __VA_ARGS__)
clock_t start_clock = clock();
#else
#define debug(...)
#define debugArr(...)
#endif
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)

{
	if (a > b)
		return a;
	return b;
}
int min(int a, int b)
{
	if (a < b)
		return a;
	return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e6 + 10;

int n, m;
char s[N];
int nxt[N];
map<ull, int> mp;

void solve()
{
	cin >> n;
	int res = 0;
	rep(i, 1, n)
	{
		cin >> m >> (s + 1);
		// cerr << m << endl;
		for (int t = 2, j = 0; t <= m; t++)
		{
			while (j && s[j + 1] != s[t])
				j = nxt[j];
			if (s[j + 1] == s[t])
				j++;
			nxt[t] = j;
		}
		int p;
		if (m % (m - nxt[m]) != 0)
			p = m;
		else
			p = m - nxt[m];
		// cerr << p << endl;
		ull hs = 0;
		rep(j, 1, p) hs = hs * 13331ull + s[j];
		res += mp[hs];
		mp[hs]++;
	}
	cout << res * 2 + n << endl;
}

#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif

signed main()
{
	// freopen(".in","r",stdin);
	// freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	while (testcase--)
		solve();
#ifndef ONLINE_JUDGE
	cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
	cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
	return 0;
}