description
solution
一开始的思路是插头\(DP\),然而复杂度太高
考虑将网格图黑白染色后跑费用流
流量为接口数，费用为操作次数
把一个方格拆成五个点，如何连边请自行脑补
打个表感觉还是挺好写的
code
#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<fstream>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<complex>
#include<clocale>
#include<cctype>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<ctime>
#include<cmath>
#include<queue>
#include<stack>
#include<list>
#include<map>
#include<set>
#define FILE "a"
#define mp make_pair
#define pb push_back
#define RG register
#define il inline
//#define RAND
using namespace std;
typedef unsigned long long ull;
typedef vector<int>VI;
typedef long long ll;
typedef double dd;
const int inf=2147483647;
const dd pi=acos(-1);
const dd eps=1e-10;
const int mod=1e9+7;
const int N=100010;
const ll INF=1e18+1;
il ll read(){
RG ll data=0,w=1;RG char ch=getchar();
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
return data*w;
}
il int make(int l,int r){return rand()%(r-l+1)+l;}
il void file(){
#ifdef RAND
freopen("seed.in","r",stdin);
RG int seed=read();fclose(stdin);
srand(time(NULL)+seed);
freopen("seed.out","w",stdout);
seed=rand();printf("%d\n",seed);
fclose(stdout);
freopen(FILE".in","w",stdout);
#endif
#ifndef RAND
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
#endif
}
/*********************************************************************/
int n,m,sum[2],p[2000][2000][4];
int w[]={0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4};
int t[]={0,1,1,3,1,2,3,4,1,3,2,4,3,4,4,5};
int dy[]={0,1,0,-1},dx[]={-1,0,1,0};
int S,T,tot,head[N],nxt[N<<1],to[N<<1],val[N<<1],cost[N<<1],cnt=1;
il void add(int u,int v,int w,int f,bool q){
if(q)swap(u,v);if(!w)return;
to[++cnt]=v;
nxt[cnt]=head[u];
val[cnt]=w;
cost[cnt]=f;
head[u]=cnt;
to[++cnt]=u;
nxt[cnt]=head[v];
val[cnt]=0;
cost[cnt]=-f;
head[v]=cnt;
}
queue<int>Q;int dis[N],pn[N],pe[N];bool vis[N];
il bool spfa(){
for(RG int i=1;i<=tot;i++)
dis[i]=inf,vis[i]=pn[i]=pe[i]=0;
while(!Q.empty())Q.pop();
Q.push(S);vis[S]=1;dis[S]=0;
while(!Q.empty()){
RG int u=Q.front();Q.pop();
for(RG int i=head[u];i;i=nxt[i]){
RG int v=to[i];
if(val[i]&&dis[v]>dis[u]+cost[i]){
dis[v]=dis[u]+cost[i];
pn[v]=u;pe[v]=i;
if(!vis[v]){Q.push(v);vis[v]=1;}
}
}
vis[u]=0;
}
return dis[T]!=inf;
}
il int solve(){
if(sum[0]!=sum[1])return -1;
RG int ans=0,ret=0,power;
while(spfa()){
power=inf;
for(RG int i=T;i!=S;i=pn[i])power=min(power,val[pe[i]]);
for(RG int i=T;i!=S;i=pn[i])
val[pe[i]]-=power,val[pe[i]^1]+=power;
ret+=power;ans+=power*dis[T];
}
return ret==sum[0]?ans:-1;
}
int main()
{
n=read();m=read();S=++tot;T=++tot;
for(RG int i=1;i<=n;i++)
for(RG int j=1;j<=m;j++){
for(RG int k=0;k<=4;k++)
p[i][j][k]=++tot;
RG int s=read(),q=((i+j)&1);sum[q]+=w[s];
add(q?T:S,p[i][j][4],w[s],0,q);
if(t[s]==1){
for(RG int k=0;k<=3;k++)
if(s&(1<<k))add(p[i][j][4],p[i][j][k],1,0,q);
else if(s&(1<<(k^2)))add(p[i][j][k^2],p[i][j][k],1,2,q);
else if(s&(1<<(k^1)))add(p[i][j][k^1],p[i][j][k],1,1,q);
else add(p[i][j][k^3],p[i][j][k],1,1,q);
}
else if(t[s]==2){
for(RG int k=0;k<=3;k++)
if(s&(1<<k))add(p[i][j][4],p[i][j][k],1,0,q);
}
else if(t[s]==3){
for(RG int k=0;k<=3;k++)
if(s&(1<<k))add(p[i][j][4],p[i][j][k],1,0,q);
else add(p[i][j][k^2],p[i][j][k],1,1,q);
}
else if(t[s]==4){
for(RG int k=0;k<=3;k++)
if(s&(1<<k))add(p[i][j][4],p[i][j][k],1,0,q);
else{
add(p[i][j][k^1],p[i][j][k],1,1,q);
add(p[i][j][k^2],p[i][j][k],1,2,q);
add(p[i][j][k^3],p[i][j][k],1,1,q);
}
}
else if(t[s]==5){
for(RG int k=0;k<=3;k++)
add(p[i][j][4],p[i][j][k],1,0,q);
}
}
for(RG int i=1;i<=n;i++)
for(RG int j=1;j<=m;j++)
if((i+j)&1)
for(RG int k=0;k<=3;k++){
RG int x=i+dx[k],y=j+dy[k];
if(x>0&&y>0&&x<=n&&y<=m)
add(p[i][j][k],p[x][y][k^2],1,0,(i+j)&1);
}
printf("%d\n",solve());
return 0;
}