P2045 方格取数加强版
傻逼题。
每个点拆成入点a和出点b，S向a(1,1)连边，b(n,n)向T连边。流量为k，费用为0.
由于点只能去一次，所以a(i,j)向b(i,j)连一条1流量-A[i][j]费用的边，再连一条k-1流量0费用的边。
再就是每个出点都向相邻入点连一条流量k费用0的边。
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#include<cstring>
#define il inline
#define rg register
#define vd void
#define sta static
typedef long long ll;
il int gi(){
rg int x=0,f=1;rg char ch=getchar();
while(ch<'0'||ch>'9')f=ch=='-'?-1:f,ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
const int maxn=50*50*2+10,maxm=1e7,S=maxn-2,T=maxn-1;
int num[51][51];
int fir[maxn],dis[maxm],nxt[maxm],w[maxm],cost[maxm],id=1;
il vd link(int a,int b,int c,int d){
nxt[++id]=fir[a],fir[a]=id,dis[id]=b,w[id]=c,cost[id]=d;
nxt[++id]=fir[b],fir[b]=id,dis[id]=a,w[id]=0,cost[id]=-d;
}
il bool SPFA(int&Cost){
sta int que[10000000],hd,tl,lst[maxn],dist[maxn];
sta bool inque[maxn]={0};
memset(dist,63,sizeof dist);
dist[S]=0;
hd=tl=0;que[tl++]=S;inque[S]=1;
while(hd^tl){
int x=que[hd];
for(rg int i=fir[x];i;i=nxt[i]){
if(w[i]&&dist[dis[i]]>dist[x]+cost[i]){
dist[dis[i]]=dist[x]+cost[i];lst[dis[i]]=i;
if(!inque[dis[i]])inque[dis[i]]=1,que[tl++]=dis[i];
}
}
inque[x]=0;++hd;
}
if(dist[T]==dist[0])return 0;
int flow=1e9;
for(rg int i=lst[T];i;i=lst[dis[i^1]])flow=std::min(flow,w[i]);
for(rg int i=lst[T];i;i=lst[dis[i^1]])w[i]-=flow,w[i^1]+=flow,Cost+=flow*cost[i];
return 1;
}
il vd Mincost(int&cst){cst=0;while(SPFA(cst));}
int main(){
#ifdef xzz
freopen("2045.in","r",stdin);
freopen("2045.out","w",stdout);
#endif
int n=gi(),k=gi();
for(rg int i=1;i<=n;++i)
for(rg int j=1;j<=n;++j)
num[i][j]=++num[0][0],++num[0][0];
for(rg int i=1;i<=n;++i)
for(rg int j=1;j<=n;++j)
link(num[i][j],num[i][j]+1,1,-gi()),link(num[i][j],num[i][j]+1,k-1,0);
link(S,num[1][1],k,0);
link(num[n][n]+1,T,k,0);
for(rg int i=1;i<=n;++i)
for(rg int j=1;j<n;++j)
link(num[i][j]+1,num[i][j+1],k,0);
for(rg int i=1;i<n;++i)
for(rg int j=1;j<=n;++j)
link(num[i][j]+1,num[i+1][j],k,0);
int Cost;Mincost(Cost);
printf("%d
",-Cost);
return 0;
}