BZOJ 1187 HNOI2007 神奇游乐园 插头DP

题目大意：给定一个四联通的网格图，每个点有个权值，求一个长度至少为4的简单环使得环上权值和最大
插头DP，具体自己讨论吧这题没啥可说的
注意$(1,2)->(0,0)$的状态要转移到答案上而不能转移到DP数组里

#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,m,ans=0xefefefef;
int a[101][7],f[102][7][1<<14];
vector<pair<int,bool> > trans[7][1<<14];
int Impress(int a[])
{
    int i,re=0;
    for(i=0;i<=m;i++)
        re|=a[i]<<i*2;
    return re;
}
void Calculate(int sta)
{
    static int a[10],b[10],another[10],stack[10];
    int i,top=0;
    for(i=0;i<=m;i++)
        a[i]=(sta>>i*2)&3;
    for(i=0;i<=m;i++)
    {
        if(a[i]==0)
            continue;
        if(a[i]==3)
            return ;
        if(a[i]==1)
            stack[++top]=i;
        else
        {
            if(top==0)
                return ;
            another[stack[top]]=i;
            another[i]=stack[top];
            stack[top--]=0;
        }
    }
    if(top) return ;
    for(i=1;i<=m;i++)
    {
        switch(a[i-1]<<2|a[i])
        {
            case 0<<2|0:
            {
                memcpy(b,a,sizeof b);
                trans[i][sta].push_back(pair<int,bool>(Impress(b),false));
                b[i-1]=1;b[i]=2;
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                break;
            }
            case 0<<2|1:
            {
                memcpy(b,a,sizeof b);
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                b[i-1]=1;b[i]=0;
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                break;
            }
            case 0<<2|2:
            {
                memcpy(b,a,sizeof b);
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                b[i-1]=2;b[i]=0;
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                break;
            }
            case 1<<2|0:
            {
                memcpy(b,a,sizeof b);
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                b[i-1]=0;b[i]=1;
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                break;
            }
            case 1<<2|1:
            {
                memcpy(b,a,sizeof b);
                b[i-1]=b[i]=0;b[another[i]]=1;
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                break;
            }
            case 1<<2|2:
            {
                memcpy(b,a,sizeof b);
                b[i-1]=b[i]=0;
                if( Impress(b)==0 )
                    trans[i][sta].push_back(pair<int,bool>(-1,true));
                break;
            }
            case 2<<2|0:
            {
                memcpy(b,a,sizeof b);
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                b[i-1]=0;b[i]=2;
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                break;
            }
            case 2<<2|1:
            {
                memcpy(b,a,sizeof b);
                b[i-1]=b[i]=0;
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                break;
            }
            case 2<<2|2:
            {
                memcpy(b,a,sizeof b);
                b[i-1]=b[i]=0;b[another[i-1]]=2;
                trans[i][sta].push_back(pair<int,bool>(Impress(b),true));
                break;
            }
        }
    }
}
int main()
{
    vector<pair<int,bool> >::iterator it;
    int i,j,k;
    cin>>n>>m;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
            scanf("%d",&a[i][j]);
    for(k=0;k<1<<(m+1<<1);k++)
        Calculate(k);
    memset(f,0xef,sizeof f);
    f[1][0][0]=0;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
            for(k=0;k<1<<(m+1<<1);k++)
                for(it=trans[j][k].begin();it!=trans[j][k].end();it++)
                {
                    if(i==1&&j==2&&k==9)
                        ++i,--i;
                    int &val=it->first!=-1?f[i][j][it->first]:ans;
                    val=max(val,f[i][j-1][k]+it->second*a[i][j]);
                }
        for(k=0;k<1<<(m<<1);k++)
            f[i+1][0][k<<2]=f[i][m][k];
    }
    cout<<ans<<endl;
    return 0;
}