bzoj2085 [Poi2010]Hamsters 矩阵快速幂+字符串hash

题目传送门

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

题解

考虑暴力 DP 的做法。令 (dp[i][j]) 表示以 (j) 为开头的子串，并且已经总共出现 (i) 次的最小长度。

[dp[i][j] = min_{k=1}^n{dp[i-1][k] + len_j - LC(j, k) } ]

其中 (LC(j, k)) 表示最长的 (j) 的后缀等于 (k) 的前缀。

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然后这个东西可以用矩阵来加速。

矩阵的定义就是先加再取 (min)。

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时间复杂度 (O(n^3log 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
", i, j, cnt);
		}
	for (int i = 1; i <= n; ++i) B.a[i][1] = len[i];//, dbg("len[%d] = %d
", 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
", i, B.a[i][1]);
	printf("%lld
", 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;
}