JZOJ 4752.字符串合成

( ext{Problem})

( ext{Solution})

最优解一定是一个回文子串的最优构造加上剩下的逐个填入
考虑用回文树建出所有的回文串，然后 (dp) 求回文子串最优的构造方案
维护一个 ( ext{half}) 意义同 ( ext{fail})，但要保证长度不超过 (len) 的一半且最大
暴力跳 ( ext{fail}) 发现不行
加上一个小 ( ext{trick})：倒着做，用 (x) 的 ( ext{half}) 更新 ( ext{fail[x]}) 的 ( ext{half})
正确性显然

( ext{Code})

#include <cstdio>
#include <cstring>
#include <iostream>
#define re register
using namespace std;

const int N = 1e5 + 5;
int T, Len;
char str[N];

struct PAM{
	int size, last, tot, tr[N][26], fail[N], half[N], len[N], f[N], fa[N];
	char s[N];
	inline int Node(int l)
	{
		++size, memset(tr[size], 0, sizeof tr[size]);
		len[size] = l, fail[size] = half[size] = 0;
		return size;
	}
	inline void clear()
	{
		size = -1, s[last = tot = 0] = '$';
		Node(0), Node(-1), fail[0] = 1;
		memset(f, 0, sizeof f), f[0] = 1;
	}
	inline int getfail(int x)
	{
		while (s[tot - len[x] - 1] != s[tot]) x = fail[x];
		return x;
	}
	inline void insert(char c)
	{
		s[++tot] = c;
		int cur = getfail(last), ch = c - 'a';
		if (!tr[cur][ch])
		{
			int u = Node(len[cur] + 2);
			fail[u] = tr[getfail(fail[cur])][ch], fa[u] = cur, tr[cur][ch] = u;
		}
		last = tr[cur][ch];
	}
	inline void gethalf()
	{
		for(re int i = 2; i <= size; i++) half[i] = fail[i];
		for(re int i = size; i >= 2; i--)
		{
			while ((len[half[i]] << 1) > len[i]) half[i] = fail[half[i]];
			if (len[half[i]] < len[half[fail[i]]]) half[fail[i]] = half[i];
		}
	}
	inline int dp()
	{
		int ans = Len;
		for(re int i = 2; i <= size; i++)
		{
			if (len[i] & 1) f[i] = min(f[fa[i]] + 2, f[fail[i]] + len[i] - len[fail[i]]);
			else f[i] = min(f[fa[i]] + 1, f[half[i]] + (len[i] >> 1) - len[half[i]] + 1);
			ans = min(ans, f[i] + Len - len[i]);
		}
		return ans;
	}
}P;

int main()
{
	scanf("%d", &T);
	for(; T; --T)
	{
		P.clear();
		scanf("%s", str + 1);
		Len = strlen(str + 1);
		for(re int i = 1; i <= Len; i++) P.insert(str[i]);
		P.gethalf();
		printf("%d
", P.dp());
	}
}