題目傳送門

https://lydsy.com/JudgeOnline/problem.php?id=2085

題解

考慮暴力 DP 的做法。令 \(dp[i][j]\) 表示以 \(j\) 為開頭的子串，并且已經(jīng)總共出現(xiàn) \(i\) 次的最小長(zhǎng)度。

其中 \(LC(j, k)\) 表示最長(zhǎng)的 \(j\) 的后綴等于 \(k\) 的前綴。

---

然后這個(gè)東西可以用矩陣來(lái)加速。

矩陣的定義就是先加再取 \(\min\)。

---

時(shí)間復(fù)雜度 \(O(n^3\log m)\)。

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)

#define dbg(...) fprintf(stderr, __VA_ARGS__)

#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)

#define fi first

#define se second

#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}

template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {

	int f = 0, c;

	while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;

	x = c & 15;

	while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);

	f ? x = -x : 0;

}

const int N = 200 + 7;

const int M = 1e5 + 7;

const ll INF = 0x3f3f3f3f3f3f3f3f;

const int base = 1997;

int n, m, mxl;

int len[N];

char s[N][M];

ull ha[N][M], bin[M];

struct Matrix {

	ll a[N][N];

	inline Matrix() { memset(a, 0x3f, sizeof(a)); }

	inline Matrix(const int &x) {

		memset(a, 0x3f, sizeof(a));

		for (int i = 1; i <= n; ++i) a[i][i] = x;

	}

	inline Matrix operator * (const Matrix &b) {

		Matrix c;

		for (int k = 1; k <= n; ++k)

			for (int i = 1; i <= n; ++i)

				for (int j = 1; j <= n; ++j)

					smin(c.a[i][j], a[i][k] + b.a[k][j]);

		return c;

	}

} A, B;

inline Matrix fpow(Matrix x, int y) {

	Matrix ans(0);

	for (; y; y >>= 1, x = x * x) if (y & 1) ans = ans * x;

	return ans;

}

inline ull get_hash(ull *h, int l, int r) { return h[r] - h[l - 1] * bin[r - l + 1]; }

inline void work() {

	bin[0] = 1;

	for (int i = 1; i <= mxl; ++i) bin[i] = bin[i - 1] * base;

	for (int i = 1; i <= n; ++i)

		for (int j = 1; j <= n; ++j) {

			int cnt = 0;

			for (int k = 1; k <= std::min(len[i], len[j]) - 1; ++k)

				if (get_hash(ha[i], 1, k) == get_hash(ha[j], len[j] - k + 1, len[j])) smax(cnt, k);

			A.a[j][i] = len[j] - cnt;

//			dbg("i = %d, j = %d, l = %d\n", i, j, cnt);

		}

	for (int i = 1; i <= n; ++i) B.a[i][1] = len[i];//, dbg("len[%d] = %d\n", i, len[i]);

	B = fpow(A, m - 1) * B;

//	B = A * B;

	ll ans = INF;

	for (int i = 1; i <= n; ++i) smin(ans, B.a[i][1]);//, dbg("B.a[%d][1] = %lld\n", i, B.a[i][1]);

	printf("%lld\n", ans);

}

inline void init() {

	read(n), read(m);

	for (int i = 1; i <= n; ++i) {

		scanf("%s", s[i] + 1);

		len[i] = strlen(s[i] + 1), smax(mxl, len[i]);

		for (int j = 1; j <= len[i]; ++j) ha[i][j] = ha[i][j - 1] * base + (s[i][j] - 'a' + 1);

	}

}

int main() {

#ifdef hzhkk

	freopen("hkk.in", "r", stdin);

#endif

	init();

	work();

	fclose(stdin), fclose(stdout);

	return 0;

}

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