洛谷 3190 「HNOI2007」神奇游乐园

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也是个板题（

使用括号表示法压缩轮廓线上的插头状态。
转移时讨论上插头和左插头的存在情况即可。

代码： 

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#include <cstdio>
#include <cstring>
#include <algorithm>
#define bits(x) ((x) - 1 << 1)
using namespace std;
const int N = 100;
const int M = 6;
const int MX = 1e6;
const int hash_mod = 70921;
int n,m,a[N + 5][M + 5];
int f[2][MX + 5],ans = -0x3f3f3f3f;
int key[2][MX + 5],tot[2],pre[2][MX + 5],first[2][hash_mod + 5];
inline void trans(int cur,int s,int v)
{
    int h = s % hash_mod;
    for(register int i = first[cur][h];i;i = pre[cur][i])
        if(key[cur][i] == s)
        {
            f[cur][i] = max(f[cur][i],v);
            return ;
        }
    key[cur][++tot[cur]] = s,f[cur][tot[cur]] = v,pre[cur][tot[cur]] = first[cur][h],first[cur][h] = tot[cur];
}
int main()
{
    scanf("%d%d",&n,&m);
    for(register int i = 1;i <= n;++i)
        for(register int j = 1;j <= m;++j)
            scanf("%d",a[i] + j);
    trans(0,0,0);
    for(register int i = 1,cur = 0;i <= n;++i)
    {
        for(register int j = 1;j <= tot[cur];++j)
            key[cur][j] <<= 2;
        for(register int j = 1;j <= m;++j,cur ^= 1)
        {
            memset(first[cur ^ 1],0,sizeof first[cur ^ 1]),tot[cur ^ 1] = 0;
            for(register int k = 1;k <= tot[cur];++k)
            {
                int s = key[cur][k],up = (s >> bits(j + 1)) & 3,left = (s >> bits(j)) & 3;
                int v = f[cur][k];
                if(!up && !left)
                    trans(cur ^ 1,s,v);
                v += a[i][j];
                if(!up && !left)
                    i < n && j < m && (trans(cur ^ 1,s + (1 << bits(j)) + (2 << bits(j + 1)),v),1);
                else if(!up && left)
                    i < n && (trans(cur ^ 1,s,v),1),
                    j < m && (trans(cur ^ 1,s - (left << bits(j)) + (left << bits(j + 1)),v),1);
                else if(up && !left)
                    j < m && (trans(cur ^ 1,s,v),1),
                    i < n && (trans(cur ^ 1,s - (up << bits(j + 1)) + (up << bits(j)),v),1);
                else if(up == 1 && left == 1)
                    for(register int l = j + 1,cnt = 0;l <= m + 1;++l)
                    {
                        ((s >> bits(l)) & 3) == 1 && ++cnt,
                        ((s >> bits(l)) & 3) == 2 && --cnt;
                        if(!cnt)
                        {
                            trans(cur ^ 1,s - (1 << bits(j)) - (1 << bits(j + 1)) - (1 << bits(l)),v);
                            break;
                        }
                    }
                else if(up == 2 && left == 2)
                    for(register int l = j,cnt = 0;l;--l)
                    {
                        ((s >> bits(l)) & 3) == 2 && ++cnt,
                        ((s >> bits(l)) & 3) == 1 && --cnt;
                        if(!cnt)
                        {
                            trans(cur ^ 1,s - (2 << bits(j)) - (2 << bits(j + 1)) + (1 << bits(l)),v);
                            break;
                        }
                    }
                else if(up == 1 && left == 2)
                    trans(cur ^ 1,s - (1 << bits(j + 1)) - (2 << bits(j)),v);
                else
                    s == (1 << bits(j)) + (2 << bits(j + 1)) && (ans = max(ans,v));
            }
        }
    }
    printf("%d\n",ans);
}