题解
平面图最小割转对偶图最短路裸题。
数组要开大！开大！开大！要特判！特判！特判！1
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using namespace std;
typedef long long ll;
int read(){
int f=1,x=0;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-f;c=getchar();}
while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
return f*x;
}
struct Edge{
int v,cost,_next;
};
Edge edge[7000005];
int cnt=0,at[2000005],n,m,S,T,sq,d[2010005];
int que[7000005];
bool in[2000005]={0};
int id(int i,int j){return (i-1)*(m-1)+j;}
void addedge(int _u,int _v,int _cost){
edge[++cnt].v=_v,
edge[cnt].cost=_cost,
edge[cnt]._next=at[_u],
at[_u]=cnt;
edge[++cnt].v=_u,
edge[cnt].cost=_cost,
edge[cnt]._next=at[_v],
at[_v]=cnt;
}
void spfa_bfs(){
fill(d+1,d+sq*2+10000,INF);
int i,_u,_v,_co,r=0,f=0,qc=0;
d[S]=0,que[r++]=S,in[S]=1,qc++;
while(qc&&qc<=7000000){
_u=que[f],in[_u]=0;
f=(f==7000000)?0:f+1;
qc--;
for(i=at[_u];i;i=edge[i]._next){
_v=edge[i].v,_co=edge[i].cost;
if(d[_u]+_co<d[_v]){
d[_v]=d[_u]+_co;
if(!in[_v]){
in[_v]=1,que[r]=_v;
r=(r==7000000)?0:r+1;
qc++;
}
}
}
}
}
void init(){
n=read(),m=read(),sq=(n-1)*(m-1);
if(n==1||m==1){
int ans=INF,c;
for(int i=1;i<=m+n-2;i++)
c=read(),ans=min(ans,c);
printf("%d\n",ans);
return ;
}
S=2*sq+1,T=2*sq+2;
int u,v,c;
for(int i=1;i<=n;i++)
for(int j=1;j<m;j++){
c=read();
if(i==1)addedge(id(i,j),T,c);
else if(i==n)addedge(id(i-1,j)+sq,S,c);
else addedge(id(i-1,j)+sq,id(i,j),c);
}
for(int i=1;i<n;i++)
for(int j=1;j<=m;j++){
c=read();
if(j==1)addedge(id(i,j)+sq,S,c);
else if(j==m)addedge(id(i,j-1),T,c);
else addedge(id(i,j-1),id(i,j)+sq,c);
}
for(int i=1;i<n;i++)
for(int j=1;j<m;j++){
c=read();
addedge(id(i,j),id(i,j)+sq,c);
}
spfa_bfs();
printf("%d\n",d[T]);
}
void solve(){
}
int main(){
init();
solve();
return 0;
}