【BZOJ 1494】【NOI 2007】生成树计数

http://www.lydsy.com/JudgeOnline/problem.php?id=1494
这道题。。因为k很小，而且我们只关心连续的k个节点的连通性，所以把连续的k个点轮廓线上的连通性的最小表示当做状态来转移。
转移可以构造一个矩阵，构造矩阵不难想，但挺麻烦的。。。
时间复杂度\(O(Mlen2^kk^2+Mlen^3logn)\)，Mlen最坏情况下为52。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned int ui;
const ui p = 65521;

ui ipow(ui a, ui b) {
	ui t = 1, w = a;
	while (b) {
		if (b & 1) t = (t * w) % p;
		w = (w * w) % p;
		b >>= 1;
	}
	return t;
}

int k;
ll n;

struct data {
	int a[5];
} state[53];

int cnt[5], tot = 0, max_now, bi[5], ref[1000003];
ui f[53], ret, rr;

void checkmin(int num) {
	rr = num;
	for (int i = 0; i < k; ++i) {
		bi[i] = num % 10;
		if (bi[i] >= k) return;
		num /= 10;
	}
	
	max_now = -1;
	for (int i = 0; i < k; ++i)
		if (bi[i] > max_now + 1) return;
		else max_now = max(max_now, bi[i]);
	
	++tot; memset(cnt, 0, sizeof(cnt)); ref[rr] = tot;
	for (int i = 0; i < k; ++i)
		state[tot].a[i] = bi[i], ++cnt[bi[i]];
	ret = 1;
	for (int i = 0; i < k; ++i)
		if (cnt[i] > 1)
			ret = ret * ipow(cnt[i], cnt[i] - 2) % p;
	f[tot] = ret;
}

struct Matrix {
	ui a[53][53];
	Matrix() {memset(a, 0, sizeof(a));}
} W, T;

Matrix operator * (Matrix A, Matrix B) {
	Matrix C;
	for (int k = 1; k <= tot; ++k)
		for (int i = 1; i <= tot; ++i)
			for (int j = 1; j <= tot; ++j)
				C.a[i][j] = (C.a[i][j] + (A.a[i][k] * B.a[k][j] % p)) % p;
	return C;
}

bool flag, vis[100003];
int col[5], ora[5];

void mark_matrix(int row, int *pre) {
	flag = true; max_now = 0;
	for (int i = 1; i < k; ++i) {
		max_now = max(max_now, pre[i]);
		if (pre[i] == pre[0])
			flag = false;
	}
	
	int totnum = (1 << (max_now + 1)), mark, lalala, now;
	ret = 0;
	for (int i = 0; i < totnum; ++i) {
		if (flag && (i & 1) == 0) continue;
		mark = -1;
		for (int j = 0; j < k; ++j) col[j] = j;
		for (int j = 0; j <= max_now; ++j)
			if ((1 << j) & i) {
				if (mark == -1) mark = j;
				else col[j] = mark;
			}
		
		memset(ora, -1, sizeof(ora)); lalala = -1;
		for (int j = 1; j < k; ++j)
			if (ora[col[pre[j]]] == -1)	bi[j - 1] = (ora[col[pre[j]]] = ++lalala);
			else bi[j - 1] = ora[col[pre[j]]];
		
		if (mark != -1 && ora[mark] != -1) bi[k - 1] = ora[mark];
		else bi[k - 1] = ++lalala;
		
		now = bi[k - 1];
		for (int j = k - 2; j >= 0; --j) now = now * 10 + bi[j];
		if (!ref[now]) continue;
		ret = 1;
		if (mark != -1)
			for (int j = 0; j < k; ++j)
				if (col[j] == mark) {
					rr = 0;
					for (int l = 0; l < k; ++l)
						if (pre[l] == j) ++rr;
					ret = ret * rr % p;
				}
		
		W.a[row][ref[now]] = ret;
	}
}

int main() {
	scanf("%d%lld", &k, &n);
	
	if (k >= n) {printf("%u\n", ipow(n, n - 2)); return 0;}
	
	int bas = 1; for (int i = 1; i <= k; ++i) bas = bas * 10;
	for (int i = 0; i <= 43210; ++i)
		if (!vis[i % bas])
			checkmin(i), vis[i % bas] = true;
	
	n -= k;
	for (int i = 1; i <= tot; ++i)
		mark_matrix(i, state[i].a);
	
/*	for (int i = 1; i <= tot; ++i) {
		for (int j = 1; j <= tot; ++j)
			printf("%d ", W.a[i][j]);
		puts("");
	}*/
	
	for (int i = 1; i <= tot; ++i)
		T.a[i][i] = 1;
	
	while (n) {
		if (n & 1) T = W * T;
		W = W * W;
		n >>= 1;
	}
	
/*	puts("");
	for (int i = 1; i <= tot; ++i) {
		for (int j = 1; j <= tot; ++j)
			printf("%d ", T.a[i][j]);
		puts("");
	}*/
	
	ui ans = 0;
	for (int i = 1; i <= tot; ++i)
		ans = (ans + f[i] * T.a[i][1] % p) % p;
	printf("%u\n", ans);
	
	return 0;
}