BZOJ3125city （插頭DP）

題意：就是上一道formula的基礎上限定某些格子只能豎着通過，某些只能橫着通過。

還是括號表示。這個轉移的時候特判一下就好了，具體實現基本和上一題一樣。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cassert>
#include<iostream>
using namespace std;
#define LL long long
#define clr(a) memset(a,0,sizeof a)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define erp(i,a,b) for(int i=a;i>=b;--i)
#define getbit(x,y) (((x) >> ((y)<<1)) & 3)
#define bit(x,y) ((x)<<((y)<<1)) 
#define clrbit(x,i,j) ((x) & (~(3<<((i)<<1))) & (~(3<<((j)<<1))))
const int MAXS = 1000000, mo = 100003;
int n, m, ex, ey;

char a[15][15];
struct Node
{
	int s, nxt;
	LL val;
};
struct Hash
{
	Node e[MAXS];
	int adj[mo], ec;
	void init()
	{
		memset(adj, -1, sizeof adj);
		ec = 0;
	}
	void push(int s, LL v)
	{
		int ha = s%mo;
		for (int i = adj[ha]; ~i; i=e[i].nxt)
			if (e[i].s == s)
			{
				e[i].val += v;
				return;
			}
		e[ec].val = v, e[ec].s = s;
		e[ec].nxt = adj[ha];
		adj[ha] = ec++;
	}
} dp[2];

inline int FindL(int st, int x)
{
	int cnt = 1, s;
	erp(i, x-1, 0)
	{
		s = (st >> (i<<1)) & 3;
		if (s == 2) cnt++;
		else if (s == 1) cnt--;
		if (!cnt) return i;
	}
	return -1;
}
inline int FindR(int st, int x)
{
	int cnt = 1, s;
	rep(i, x+1, m)
	{
		s = (st >> (i<<1)) & 3;
		if (s == 1) cnt++;
		else if (s == 2) cnt--;
		if (!cnt) return i;
	}
	return -1;
}

void work1(int i, int j, int cur) // '.'
{
	dp[cur].init();
	rep(k, 0, dp[cur^1].ec-1)
	{
		int L = getbit(dp[cur^1].e[k].s, j-1);
		int U = getbit(dp[cur^1].e[k].s, j);
		int s = clrbit(dp[cur^1].e[k].s, j-1, j);
		LL las = dp[cur^1].e[k].val;
		if (!L && !U)
		{
			if (i<n && j<m)
				dp[cur].push(s|bit(1, j-1)|bit(2, j), las);
		}
		else if (!L)
		{
			if (i < n) dp[cur].push(s|bit(U, j-1), las);
			if (j < m) dp[cur].push(s|bit(U, j), las);
		}
		else if (!U)
		{
			if (i < n) dp[cur].push(s|bit(L, j-1), las);
			if (j < m) dp[cur].push(s|bit(L, j), las);
		}
		else if (L==1 && U==1)
			dp[cur].push(s^bit(3, FindR(s, j)), las);
		else if (L==2 && U==2)
			dp[cur].push(s^bit(3, FindL(s, j-1)), las);
		else if (L==2 && U==1)
			dp[cur].push(s, las);
		else if (i==ex && j==ey)
			dp[cur].push(s, las);
	}
}

void work2(int i, int j, int cur)
{
	dp[cur].init();
	rep(k, 0, dp[cur^1].ec-1)
	{
		int L = getbit(dp[cur^1].e[k].s, j-1);
		int U = getbit(dp[cur^1].e[k].s, j);
		int s = clrbit(dp[cur^1].e[k].s, j-1, j);
		LL las = dp[cur^1].e[k].val;
		if (!L && !U) dp[cur].push(s, las);
	}
}

void worklr(int i, int j, int cur)
{
	dp[cur].init();
	rep(k, 0, dp[cur^1].ec-1)
	{
		int L = getbit(dp[cur^1].e[k].s, j-1);
		int U = getbit(dp[cur^1].e[k].s, j);
		int s = clrbit(dp[cur^1].e[k].s, j-1, j);
		LL las = dp[cur^1].e[k].val;
		if (!U && L)
		{
			if (j < m) dp[cur].push(s|bit(L, j), las);
		}
	}
}

void workud(int i, int j, int cur)
{
	dp[cur].init();
	rep(k, 0, dp[cur^1].ec-1)
	{
		int L = getbit(dp[cur^1].e[k].s, j-1);
		int U = getbit(dp[cur^1].e[k].s, j);
		int s = clrbit(dp[cur^1].e[k].s, j-1, j);
		LL las = dp[cur^1].e[k].val;
		if (!L && U)
		{
			if (i < n) dp[cur].push(s|bit(U, j-1), las);
		}
	}
}

LL solve()
{
	dp[0].init(), dp[0].push(0, 1);
	int cur = 0;
	rep(i, 1, n)
	{
		rep(k, 0, dp[cur].ec-1)
			dp[cur].e[k].s <<= 2;
		rep(j, 1, m)
		{
			cur ^= 1;
			if (a[i][j]=='.') work1(i, j, cur);
			else if (a[i][j]=='#') work2(i, j, cur);
			else if (a[i][j]=='-') worklr(i, j, cur);
			else workud(i, j, cur);
		}
	}
	rep(k, 0, dp[cur].ec-1)
		if (!dp[cur].e[k].s) return dp[cur].e[k].val;
	return 0;
}

int main()
{
	scanf("%d%d", &n, &m);
	rep(i, 1, n)
	{
		scanf("%s", a[i]+1);
		rep(j, 1, m)
		{
			if (a[i][j]=='.') ex = i, ey = j;
			if (a[i][j]=='-'&&(j==1||j==m)) { puts("0"); return 0; }
			if (a[i][j]=='|'&&(i==1||i==n)) { puts("0"); return 0; }
		}
	}
	cout << solve() << '\n';
	return 0;
}