ural1519 Formula 1

Background
Regardless of the fact, that Vologda could not get rights to hold the Winter Olympic games of 20**, it is well-known, that the city will
conduct one of the Formula 1 events. Surely, for such an important
thing a new race circuit should be built as well as hotels,
restaurants, international airport - everything for Formula 1 fans,
who will flood the city soon. But when all the hotels and a half of
the restaurants were built, it appeared, that at the site for the
future circuit a lot of gophers lived in their holes. Since we like
animals very much, ecologists will never allow to build the race
circuit over the holes. So now the mayor is sitting sadly in his
office and looking at the map of the circuit with all the holes
plotted on it.
Problem
Who will be smart enough to draw a plan of the circuit and keep the city from inevitable disgrace? Of course, only true professionals
- battle-hardened programmers from the first team of local technical university!.. But our heroes were not looking for easy life and set
much more difficult problem: “Certainly, our mayor will be glad, if we
find how many ways of building the circuit are there!” - they said.
It should be said, that the circuit in Vologda is going to be rather simple. It will be a rectangle N* M cells in size with a single
circuit segment built through each cell. Each segment should be
parallel to one of rectangle’s sides, so only right-angled bends may
be on the circuit. At the picture below two samples are given for N =
M = 4 (gray squares mean gopher holes, and the bold black line means
the race circuit). There are no other ways to build the circuit here.
Problem illustration Input
The first line contains the integer numbers N and M (2 ≤ N, M ≤ 12). Each of the next N lines contains M characters, which are the
corresponding cells of the rectangle. Character “.” (full stop) means
a cell, where a segment of the race circuit should be built, and
character “*” (asterisk) - a cell, where a gopher hole is located.
There are at least 4 cells without gopher holes. Output
You should output the desired number of ways. It is guaranteed, that it does not exceed 2 63-1.

插头dp，括号序列维护连通性。

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;#define LL long longconst int p=233333;int clo,n,m,i,j;LL ans;char map[15][15];struct res{    int fir[240010],ne[2000010],val[2000010],last[240010],tot,k,l;    LL ans[2000010];    void init()    {        tot=0;    }    int find(int x)    {        int y=x%p;        if (last[y]<clo)        {            last[y]=clo;            fir[y]=0;        }        for (int i=fir[y];i;i=ne[i])            if (val[i]==x) return i;        ne[++tot]=fir[y];        fir[y]=tot;        val[tot]=x;        ans[tot]=0;        return tot;    }    void upd(int x,LL y)    {        if (j==m-1)        {            if ((x>>2*m)&3) return;            x<<=2;        }        int z=find(x);        ans[z]+=y;    }    int next(LL &x)    {        if (ne[l])        {            l=ne[l];            x=ans[l];            return val[l];        }        for (k++;k<p;k++)        if (last[k]==clo-1)            for (l=fir[k];l;l=ne[l])            {                x=ans[l];                return val[l];            }        return -1;    }    int first(LL &x)    {        for (k=0;k<p;k++)        if (last[k]==clo-1)            for (l=fir[k];l;l=ne[l])            {                x=ans[l];                return val[l];            }    }}dp[2];int find_next(int x,int p){    int cnt=0,tem;    for (int i=p+1;;i++)    {        tem=(x>>2*i)&3;        if (tem==1) cnt++;        if (tem==2)        {            if (!cnt) return i;            cnt--;        }    }}int find_pre(int x,int p){    int cnt=0,tem;    for (int i=p-1;;i--)    {        tem=(x>>2*i)&3;        if (tem==1)        {            if (!cnt) return i;            cnt--;        }        if (tem==2) cnt++;    }}void pause(){    int i;    i=1;}int main(){    int lx,ly,x,x1,w,p,q,flag;    LL v;    scanf("%d%d",&n,&m);    for (i=0;i<n;i++)        scanf("%s",map[i]);    if (n<m)    {        for (int i=0;i<m;i++)            for (int j=i+1;j<m;j++)                swap(map[i][j],map[j][i]);        swap(n,m);    }    flag=1;    for (i=n-1;i>=0&&flag;i--)        for (j=m-1;j>=0&&flag;j--)            if (map[i][j]=='.')            {                lx=i;                ly=j;                flag=0;            }    dp[0].init();    dp[0].upd(0,1);    clo=0;    for (i=0;i<n;i++)        for (j=0;j<m;j++)        {            clo++;            dp[clo&1].init();            //if (i==3&&j==1) pause();            for (x=dp[clo&1^1].first(v);x>=0;x=dp[clo&1^1].next(v))            {                p=(x>>2*j)&3;                q=(x>>2*(j+1))&3;                x1=x^(p<<2*j)^(q<<2*(j+1));                if (map[i][j]=='*')                {                    if (p+q==0) dp[clo&1].upd(x,v);                    continue;                }                if (p+q==0) dp[clo&1].upd(x|(1<<2*j)|(2<<2*(j+1)),v);                else                {                    if (p*q==0)                    {                        dp[clo&1].upd(x1|(p+q<<2*j),v);                        dp[clo&1].upd(x1|(p+q<<2*(j+1)),v);                    }                    else                    {                        if (p==1)                        {                            if (q==1)                            {                                w=find_next(x,j+1);                                dp[clo&1].upd(x1^(3<<2*w),v);                            }                            else                            {                                if (i==lx&&j==ly&&!x1) ans+=v;                            }                        }                        else                        {                            if (q==1) dp[clo&1].upd(x1,v);                            else                            {                                w=find_pre(x,j);                                dp[clo&1].upd(x1^(3<<2*w),v);                            }                        }                    }                }            }        }    cout<<ans<<endl;}