【题目链接】
https://www.lydsy.com/JudgeOnline/problem.php?id=2310
【题解】
插头dp中一道较为繁琐（？）的题。
多开一维状态记选了多少个度为1的点。同时在状态的括号序列中新添一种状态3表示单独的插头，接下来就是分类讨论了。细节有点多，但是思维难度不大,具体实现见代码。
时间复杂度O(N∗M∗2(N+2)∗2)O(N∗M∗2(N+2)∗2)
/* --------------
user Vanisher
problem bzoj-2310
----------------*/
# include
# define ll long long
# define inf 0x3f3f3f3f
# define N 110
# define T 10
using namespace std;
int read(){
int tmp=0, fh=1; char ch=getchar();
while (ch<'0'||ch>'9'){if (ch=='-') fh=-1; ch=getchar();}
while (ch>='0'&&ch<='9'){tmp=tmp*10+ch-'0'; ch=getchar();}
return tmp*fh;
}
int now[T],nex[T],n,m,mp[N][N],ne[N][N],f1,f2,lim,h[1<<(T*2)],ans;
vector <int> g[2],f[2];
int findopp(int id){
if (now[id]==1){
int p=id+1, cnt=0;
while (cnt!=0||now[p]!=2){
if (now[p]==1) cnt++;
if (now[p]==2) cnt--;
p++;
}
return p;
}
if (now[id]==2){
int p=id-1, cnt=0;
while (cnt!=0||now[p]!=1){
if (now[p]==1) cnt++;
if (now[p]==2) cnt--;
p--;
}
return p;
}
return -1;
}
int getnum(){
int num=0;
for (int i=0; i<=m+1; i++)
num=(nex[i]<<(i*2))+num;
return num;
}
void join(int num){
if (nex[0]>2) return;
int tmp=getnum(),cnt=0;
for (int i=1; i<=m+1; i++){
if (nex[i]==1) cnt++;
if (nex[i]==2) cnt--;
}
if (h[tmp]==-1){
h[tmp]=g[f2].size();
g[f2].push_back(tmp);
f[f2].push_back(num);
}
else f[f2][h[tmp]]=max(f[f2][h[tmp]],num);
}
int main(){
n=read(), m=read();
for (int i=1; i<=n; i++)
for (int j=1; j<=m; j++)
mp[i][j]=read();
if (nfor (int i=1; i<=n; i++)
for (int j=1; j<=m; j++)
ne[j][i]=mp[i][j];
swap(n,m);
for (int i=1; i<=n; i++)
for (int j=1; j<=m; j++)
mp[i][j]=ne[i][j];
}
f1=0, f2=1, lim=1<<((m+2)*2);
g[f1].push_back(0);
f[f1].push_back(0);
memset(h,-1,sizeof(h));
for (int i=1; i<=n; i++){
for (int j=1; j<=m; j++){
for (unsigned k=0; kint tmp=g[f1][k], num=f[f1][k], ths=mp[i][j];
if (tmp>=lim) continue;
for (int t=0; t<=m+1; t++) now[t]=nex[t]=(tmp>>(t*2))&3;
if (now[j]==3){
if (now[j+1]==0){
nex[0]=now[0]+0, nex[j]=3, nex[j+1]=0; join(num+ths);
nex[0]=now[0]+0, nex[j]=0, nex[j+1]=3; join(num+ths);
nex[0]=now[0]+1, nex[j]=0, nex[j+1]=0; join(num+ths);
}
else if (now[j+1]==3){
nex[0]=now[0]+0, nex[j]=0, nex[j+1]=0; join(num+ths);
}
else {
int k=findopp(j+1);
if (k!=-1){nex[j]=0; nex[j+1]=0; nex[k]=3; join(num+ths);}
}
continue;
}
if (now[j+1]==3){
if (now[j]==0){
nex[0]=now[0]+0, nex[j]=3, nex[j+1]=0; join(num+ths);
nex[0]=now[0]+0, nex[j]=0, nex[j+1]=3; join(num+ths);
nex[0]=now[0]+1, nex[j]=0, nex[j+1]=0; join(num+ths);
}
else if (now[j]==3){
nex[0]=now[0]+0, nex[j]=0, nex[j+1]=0; join(num+ths);
}
else {
int k=findopp(j);
if (k!=-1){nex[j]=0; nex[j+1]=0; nex[k]=3; join(num+ths);}
}
continue;
}
if (now[j]!=0&&now[j+1]!=0){
if (now[j]==2&&now[j+1]==1){
nex[j]=0, nex[j+1]=0; join(num+ths);
}
if (now[j]==1&&now[j+1]==1){
int p=findopp(j+1); nex[j]=0, nex[j+1]=0, nex[p]=1; join(num+ths);
}
if (now[j]==2&&now[j+1]==2){
int p=findopp(j); nex[j]=0, nex[j+1]=0, nex[p]=2; join(num+ths);
}
continue;
}
if (now[j]==0&&now[j+1]==0){
nex[j]=0, nex[j+1]=0; join(num);
nex[j]=1, nex[j+1]=2; join(num+ths);
nex[0]=now[0]+1, nex[j]=3, nex[j+1]=0; join(num+ths);
nex[0]=now[0]+1, nex[j+1]=3, nex[j]=0; join(num+ths);
nex[0]=now[0]+2, nex[j]=0, nex[j+1]=0; join(num+ths);
continue;
}
if (now[j]!=0){
nex[j]=0, nex[j+1]=now[j]; join(num+ths);
nex[j]=now[j], nex[j+1]=0; join(num+ths);
int p=findopp(j);
nex[0]=now[0]+1, nex[j]=0, nex[j+1]=0; nex[p]=3; join(num+ths);
continue;
}
if (now[j+1]!=0){
nex[j]=0, nex[j+1]=now[j+1]; join(num+ths);
nex[j]=now[j+1], nex[j+1]=0; join(num+ths);
int p=findopp(j+1);
nex[0]=now[0]+1, nex[j]=0, nex[j+1]=0; nex[p]=3; join(num+ths);
continue;
}
}
for (unsigned k=0; k1;
f[f1].clear(), g[f1].clear();
swap(f1,f2);
}
for (unsigned k=0; kint tmp=g[f1][k]&3; g[f1][k]=(g[f1][k]>>2)<<2;
g[f1][k]=(g[f1][k]<<2)+tmp;
}
}
for (unsigned i=0; iif (g[f1][i]==2){
ans=f[f1][i];
break;
}
printf("%d\n",ans);
return 0;
}