插头DP--URAL1519Formula 1

去年的老朋友。挺怀念的，回来看看。

$n \leq 12,m \leq 12$，$n*m$的01矩形，问在0中走路的欧拉回路数。答案保证longlong范围。

先设计插头：左右括号和空插头；然后分3*3种情况转移。耐心。

 //#include<iostream>
 #include<cstring>
 #include<cstdio>
 //#include<time.h>
 //#include<complex>
 #include<algorithm>
 #include<stdlib.h>
 using namespace std;

 int n,m; bool mp[][];
 #define maxn 2333333
 #define LL long long
 int state[][maxn],len[]; LL ans[][maxn]; int cur;
 int Next[maxn],first[maxn];

 void insert(int x,LL v)
 {
     int y=cur^,h=x%maxn;
     for (int i=first[h];i;i=Next[i]) if (state[y][i]==x) {ans[y][i]+=v; return;}
     state[y][++len[y]]=x; ans[y][len[y]]=v;
     Next[len[y]]=first[h]; first[h]=len[y];
 }

 int main()
 {
     scanf("%d%d",&n,&m);
     for (int i=;i<=n;i++)
         for (int j=;j<=m;j++)
         {
             char c; while ((c=getchar())!='.' && c!='*');
             mp[i][j]=(c=='*');
         }
     //0 kong di 1 zhang ai

     int endx=,endy;
     for (int i=n;i && !endx;i--)
         for (int j=m;j;j--)
             if (!mp[i][j]) {endx=i; endy=j; break;}

     cur=; insert(,); cur=; LL Ans=;
     for (int i=;i<=n;i++)
         for (int j=;j<=m;j++)
         {
             for (int k=;k<=len[cur];k++) first[state[cur][k]%maxn]=;
             for (int k=;k<=len[cur];k++)
             {
                 int now=state[cur][k],p=now&,q=(now>>)&,nk;
                 if (mp[i][j]) {if (p== && q==) nk=now>>,insert(nk,ans[cur][k]);}
                 else if (p== && q==) {if (j<m) {nk=(now>>)||(<<(m<<)); insert(nk,ans[cur][k]);}}
                 else if (p== && q)
                 {
                     nk=((now>>)^q)|(q<<(m<<)); insert(nk,ans[cur][k]);
                     if (j<m) {nk=now>>; insert(nk,ans[cur][k]);}
                 }
                 else if (p && q==)
                 {
                     nk=(now>>)|(p<<(m<<)); insert(nk,ans[cur][k]);
                     if (j<m) {nk=(now>>)|p; insert(nk,ans[cur][k]);}
                 }
                 else if (p== && q==)
                 {
                     int cnt=; nk=(now>>)^q;
                     for (int l=;;l++)
                     {
                         int hh=(now>>(l<<))&;
                         if (hh==) cnt++; if (hh==) cnt--;
                         if (cnt==) {nk=(nk^(<<((l-)<<)))^(<<((l-)<<)); break;}
                     }
                     insert(nk,ans[cur][k]);
                 }
                 else if (p== && q==)
                 {
                     nk=(now>>)^q;
                     insert(nk,ans[cur][k]);
                 }
                 else if (p== && q==) {if (i==endx && j==endy && (now^p^(q<<))==) Ans+=ans[cur][k];}
                 else if (p== && q==)
                 {
                     int cnt=,nk=(now>>)^q;
                     for (int l=m;;l--)
                     {
                         int hh=(now>>(l<<))&;
                         if (hh==) cnt++; if (hh==) cnt--;
                         if (cnt==) {nk=(nk^(<<((l-)<<)))^(<<((l-)<<)); break;}
                     }
                     insert(nk,ans[cur][k]);
                 }
             }
             len[cur]=;
             cur^=;
         }
     printf("%lld\n",Ans);
     return ;
 }