【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;
}