codeforces gym 100548G

求两个字符串的公共回文子串个数，搞两个回文树，dfs一遍即可。

#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 mod = 1e9 + 7;
const int maxn = 2e5 + 10;
LL ans;
int T;
char s[maxn];

struct PalindromicTree
{
	const static int maxn = 2e5 + 10;
	const static int size = 26;
	int next[maxn][size], sz, tot;
	int fail[maxn], len[maxn], last;
	LL cnt[maxn];
	char s[maxn];
	LL operator[](const int &x) { return cnt[x]; }
	void clear()
	{
		len[1] = -1; len[2] = 0;
		fail[1] = fail[2] = 1;
		cnt[1] = cnt[2] = tot = 0;
		last = (sz = 3) - 1;	
		memset(next[1], 0, sizeof(next[1]));
		memset(next[2], 0, sizeof(next[2]));
	}
	int Node(int length)
	{
		memset(next[sz], 0, sizeof(next[sz]));
		len[sz] = length;  cnt[sz] = 0;  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 (next[y][x]) { last = next[y][x]; }
		else {
			last = next[y][x] = Node(len[y] + 2);
			fail[last] = len[last] == 1 ? 2 : next[getfail(fail[y])][x];
		}
		return ++cnt[last];
	}
	void work()
	{
		for (int i = sz - 1; i > 2; i--)
		{
			if (fail[i] > 2) cnt[fail[i]] += cnt[i];
		}
	}
}work[2];

void dfs(int x, int y)
{
	ans += work[0][x] * work[1][y];
	for (int i = 0; i < 26; i++)
	{
		if (work[0].next[x][i] && work[1].next[y][i])
		{
			dfs(work[0].next[x][i], work[1].next[y][i]);
		}
	}
}

int main()
{
	scanf("%d", &T);
	for (int cas = 1; cas <= T; cas++)
	{
		for (int i = 0; i < 2; i++)
		{
			work[i].clear();
			scanf("%s", s);
			for (int j = 0; s[j]; j++)
			{
				work[i].add(s[j]);
			}
			work[i].work();
		}
		ans = 0;
		dfs(1, 1);
		dfs(2, 2);
		printf("Case #%d: %lld\n", cas, ans);
	}
	return 0;
}