【URAL 1519】Formula 1

http://acm.timus.ru/problem.aspx?space=1&num=1519
调了好久啊。参考（抄）的iwtwiioi的题解。
如果想要题解，题解在《基于连通性状态压缩的动态规划问题》。
我肯定讲得不如题解清楚QAQ，所以还是不讲了(╯‵□′)╯︵┻━┻
连通性状压得用bfs扩展同时hash表判重。为什么呢？
因为状态太多数组存不下，只能用hash表存有用的状态，没错就是这样！！！
感觉代码量比普通状压高了不少QAQ
这么复杂，还是典型例题→_→。
时间复杂度\(O(n*m^22^{(m+1)*2})\)。
我用的四进制表示状态，常数比三进制小。bfs和hash表应该能跑得比理论复杂度快得多吧。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define BIT(a, b) ((a) << ((b) << 1))
#define CLR(a, b) (a ^= ((a) & BIT(3, b)))
#define GET(a, b) (((a) >> ((b) << 1)) & 3)
using namespace std;
const int M = 1000007;
typedef long long ll;

bool mp[12][12], flag;

ll ans = 0;

int n, m, lan = 0, lam = 0;

struct node {
	int to[M], hash[M], cnt;
	ll sum[M];
	
	node() {
		memset(to, 0, sizeof(to));
		memset(hash, -1, sizeof(hash));
		memset(sum, 0, sizeof(sum));
		cnt = 0;
	}
	
	void ins(int s, ll num) {
		int pos = s % M; flag = false;
		while (true) {
			if (hash[pos] == -1) {hash[pos] = s; flag = true; break;}
			if (hash[pos] == s) break;
			++pos; if (pos == M) pos = 0;
		}
		sum[pos] += num;
		if (flag) to[++cnt] = pos;
	}
	
	void clr() {while (cnt) {hash[to[cnt]] = -1; sum[to[cnt]] = 0; --cnt;}}	
} T1, T2;

int cc = 0, nn;

int find(int s, int tmp, int d) {
	if (d) {
		for (int i = tmp; i < m; ++i) {
			nn = GET(s, i);
			if (nn == 1) ++cc;
			if (nn == 2) --cc;
			if (cc == 0) return i;
		}
	} else {
		for (int i = tmp; i >= 0; --i) {
			nn = GET(s, i);
			if (nn == 2) ++cc;
			if (nn == 1) --cc;
			if (cc == 0) return i;
		}
	}
}

int l, u, d, r, pos;

bool getnxt(int s, int row, int col, bool U, bool D, bool L, bool R, int &t, ll sum) {
	if ((row == 0 && U) || (col == 0 && L) || (row == n - 1 && D) || (col == m - 1 && R)) return false;
	if ((mp[row + 1][col] && D) || (mp[row][col + 1] && R)) return false;
	if (row == lan && col == lam && (R || D)) return false;
	l = GET(s, col); u = GET(s, col + 1);
	if ((((bool) u) != U) || (((bool) l) != L)) return false;
	t = s; CLR(t, col + 1); CLR(t, col);
	
	d = r = 0;
	if (l == 0 && u == 0) {
		if (D && R)
			d = 1, r = 2;
	} else if (l && u) {
		if (l == 1 && u == 2) {
			if (row != lan || col != lam) return false;
			ans += sum;
		} else if (l == 1 && u == 1) {
			pos = find(s, col + 1, 1);
			CLR(t, pos);
			t |= BIT(1, pos);
		} else if (l == 2 && u == 2) {
			pos = find(s, col, 0);
			CLR(t, pos);
			t |= BIT(2, pos);
		}
	} else if (l && !u) {
		if (D) d = l, r = 0;
		if (R) r = l, d = 0;
	} else if (!l && u) {
		if (D) d = u, r = 0;
		if (R) r = u, d = 0;
	}
	
	t |= BIT(d, col); t |= BIT(r, col + 1);
	if (col == m - 1) t <<= 2;
	return true;
}

void bfs() {
	node *q1, *q2; ll sum; int s, t;
	q1 = &T1; q2 = &T2;
	q1->ins(0, 1);
	for (int row = 0; row < n; ++row)
		for (int col = 0; col < m; ++col) {
			q2->clr();
			for (int i = q1->cnt; i >= 1; --i) {
				sum = q1->sum[q1->to[i]];
				s = q1->hash[q1->to[i]];
				if (mp[row][col]) {
					if (getnxt(s, row, col, 0, 0, 0, 0, t, sum))
						q2->ins(t, sum);
				} else {
					if (getnxt(s, row, col, 1, 1, 0, 0, t, sum)) q2->ins(t, sum);
					if (getnxt(s, row, col, 1, 0, 1, 0, t, sum)) q2->ins(t, sum);
					if (getnxt(s, row, col, 1, 0, 0, 1, t, sum)) q2->ins(t, sum);
					if (getnxt(s, row, col, 0, 1, 1, 0, t, sum)) q2->ins(t, sum);
					if (getnxt(s, row, col, 0, 1, 0, 1, t, sum)) q2->ins(t, sum);
					if (getnxt(s, row, col, 0, 0, 1, 1, t, sum)) q2->ins(t, sum);
				}
			}
			swap(q1, q2);
			if (row == lan && col == lam) return;
		}
}

int main() {
	scanf("%d%d", &n, &m);
	char c;
	for (int i = 0; i < n; ++i)	
		for (int j = 0; j < m; ++j) {
			for (c = getchar(); c != '*' && c != '.'; c = getchar());
			if (c == '*') mp[i][j] = true;
			else lan = i, lam = j;
		}
	
	bfs();
	printf("%lld\n", ans);
	return 0;
}