[CERC2014]Virus synthesis：

题目大意：

初始有一个空串，利用下面的操作构造给定串 \(S\) 。

1、串开头或末尾加一个字符

2、串开头或末尾加一个该串的逆串

求最小化操作数， \(|S| \leq 10^5\) 。

思路：

回文自动机好题。在求出失配指针时，求出 \(\mathrm{trans}_i\) 表示以 \(i\) 结尾的回文串后半段起始位置。

然后可以在回文树上 DP，设 \(f_i\) 表示变成 \(i\) 串的最小代价。

那么有：

\[\begin{aligned} f_u&\leftarrow f_{\mathrm{trans}_u} + 1 + \frac{\mathrm{len}_u}{2} - \mathrm{len}_{\mathrm{trans}_u}\\ f_u&\leftarrow f_{\mathrm{fa}_u}+1 \end{aligned} \]

那么 \(u\) 串的贡献为 \(n+f_u-\mathrm{len}_u\)

代码：

const int N = 1e5 + 10;

inline ll Read()
{
	ll x = 0, f = 1;
	char c = getchar();
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') f = -f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
	return x * f;
}

int t;
char s[N];

int Code(char c)
{
	switch (c)
	{
		case 'A': return 1;
		case 'G': return 2;
		case 'C': return 3;
		case 'T': return 4;
	}
}

AK PAM
{
	int s[N], len[N], trans[N], fail[N], t[N][26];
	int tot, lst, n;
	int f[N], ans;
	int New(int x)
	{
		len[++tot] = x;
		return tot;
	}
	void Build()
	{
		for (int i = 0; i <= tot; i++)
			for (int j = 1; j <= 4; j++)
				t[i][j] = 0;
		tot = -1, n = 0;
		New(0), New(-1);
		s[0] = -1, fail[0] = 1;
	}
	
	int Find(int x)
	{
		while (s[n - 1 - len[x]] != s[n]) x = fail[x];
		return x;
	}
	
	void Insert(int x)
	{
		s[++n] = x;
		int cur = Find(lst);
		if (!t[cur][x])
		{
			int now = New(len[cur] + 2),
			tmp = Find(fail[cur]);
			fail[now] = t[tmp][x];
			t[cur][x] = now;
			if (len[now] <= 2) trans[now] = fail[now];
			else
			{
				tmp = trans[cur];
				while (s[n - 1 - len[tmp]] != s[n] || (len[tmp] + 2 << 1) > len[now]) tmp = fail[tmp];
				trans[now] = t[tmp][x];
			}
		}
		lst = t[cur][x];
	}
	void Solve(int n)
	{
		queue<int> q;
		for (int i = 1; i <= 4; i++)
			if (t[0][i]) q.push(t[0][i]);
		for (int i = 2; i <= tot; i++) f[i] = len[i];
		while (!q.empty())
		{
			int u = q.front(); q.pop();
			f[u] = min(f[u], f[trans[u]] + 1 + len[u] / 2 - len[trans[u]]);
			ans = min(ans, n - len[u] + f[u]);
			for (int i = 1; i <= 4; i++)
				if (t[u][i])
				{
					int v = t[u][i];
					f[v] = min(f[v], f[u] + 1);
					q.push(v);
				}
		}
	}
}


int main()
{
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	for (t = Read(); t--; )
	{
		scanf ("%s", s + 1);
		PAM::Build();
		int len = strlen(s + 1);
		for (int i = 1; i <= len; i++)
			PAM::Insert(Code(s[i]));
		PAM::ans = len;
		PAM::Solve(len); 
		printf ("%d\n", PAM::ans);
	}
	return 0;
}

标签：tmp,P4762,int,Luogu,len,getchar,CERC2014,trans,mathrm
来源： https://www.cnblogs.com/GJY-JURUO/p/15152669.html