题目地址：POJ 3422

方法是对每个点i拆点成i‘和i‘‘，然后对每个i‘和i‘‘连一条费用为该点值，流量为1的边，再连1条费用为0，流量为k-1的边。

然后对每个点与右边下边相邻的点连边，流量均为INF，费用均为0。需要再建一源点与汇点，对于k次只需要在源点与汇点处进行限制即可。

代码如下：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <queue>
#include <map>
#include <set>
#include <algorithm>

using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=6000;
int head[maxn], source, sink, cnt, flow, cost, n;
int d[maxn], vis[maxn], cur[maxn], f[maxn];
int mp[100][100];
struct node
{
    int u, v, cap, cost, next;
} edge[1000000];
void add(int u, int v, int cap, int cost)
{
    edge[cnt].v=v;
    edge[cnt].cap=cap;
    edge[cnt].cost=cost;
    edge[cnt].next=head[u];
    head[u]=cnt++;

    edge[cnt].v=u;
    edge[cnt].cap=0;
    edge[cnt].cost=-cost;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}
int spfa()
{
    memset(vis,0,sizeof(vis));
    memset(d,INF,sizeof(d));
    deque<int>q;
    q.push_back(source);
    d[source]=0;
    cur[source]=-1;
    f[source]=INF;
    while(!q.empty())
    {
        int u=q.front();
        q.pop_front();
        vis[u]=0;
        for(int i=head[u]; i!=-1; i=edge[i].next)
        {
            int v=edge[i].v;
            if(d[v]>d[u]+edge[i].cost&&edge[i].cap)
            {
                d[v]=d[u]+edge[i].cost;
                f[v]=min(f[u],edge[i].cap);
                cur[v]=i;
                if(!vis[v])
                {
                    if(!q.empty()&&d[v]<d[q.front()])
                    {
                        q.push_front(v);
                    }
                    else
                        q.push_back(v);
                    vis[v]=1;
                }
            }
        }
    }
    if(d[sink]==INF) return 0;
    flow+=f[sink];
    cost-=d[sink]*f[sink];
    for(int i=cur[sink]; i!=-1; i=cur[edge[i^1].v])
    {
        edge[i].cap-=f[sink];
        edge[i^1].cap+=f[sink];
    }
    return 1;
}
void mcmf()
{
    flow=cost=0;
    while(spfa()) ;
    printf("%d\n",cost);
}
int main()
{
    int i, j, k;
    scanf("%d%d",&n,&k);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            scanf("%d",&mp[i][j]);
        }
    }
    memset(head,-1,sizeof(head));
    cnt=0;
    source=0;
    sink=2*n*n+1;
    add(source,1,k,0);
    add(2*n*n,sink,k,0);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            add((i-1)*n+j,(i-1)*n+j+n*n,1,-mp[i][j]);
            add((i-1)*n+j,(i-1)*n+j+n*n,k-1,0);
            if(i+1<=n)
            {
                add((i-1)*n+j+n*n,i*n+j,k,0);
            }
            if(j+1<=n)
            {
                add((i-1)*n+j+n*n,(i-1)*n+j+1,k,0);
            }
        }
    }
    mcmf();
    return 0;
}