题目链接：

pid=2686">http://acm.hdu.edu.cn/showproblem.php?pid=2686

和POJ3422一样

删掉K把汇点与源点的容量改为2（由于有两个方向的选择）就可以

#include <iostream>

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <queue>

#include <algorithm>

const int maxn = 20000;

const int maxm = 800000;

const int inf = 1e8;

const int INF = 0x3f3f3f3f;

#define MIN INT_MIN

#define MAX 1e6

#define LL long long

#define init(a) memset(a,0,sizeof(a))

#define FOR(i,a,b) for(int i = a;i<b;i++)

#define max(a,b) (a>b)?(a):(b)

#define min(a,b) (a>b)?(b):(a)

using namespace std;

struct node

{

    int u,v,w,cap,next;

} edge[maxm];

int pre[maxn],dis[maxn],head[maxn],cnt;

bool vis[maxn];

int n;

void add(int u,int v,int c,int cap)

{

    edge[cnt].u=u;

    edge[cnt].v=v;

    edge[cnt].w=c;

    edge[cnt].cap=cap;

    edge[cnt].next=head[u];

    head[u]=cnt++;

    edge[cnt].u=v;

    edge[cnt].v=u;

    edge[cnt].w=-c;

    edge[cnt].cap=0;

    edge[cnt].next=head[v];

    head[v]=cnt++;

}

int spfa(int s,int t)

{

    queue<int>q;

    while(q.empty()==false) q.pop();

    q.push(s);

    memset(vis,0,sizeof(vis));

    memset(pre,-1,sizeof(pre));

    FOR(i,s,t+1)

    dis[i] = -1;//求最长路dis数组初始化为-1

    dis[s]=0;

    vis[s] = 1;

    while(!q.empty())

    {

        int u=q.front();

        q.pop();

        vis[u] = 0;

        for(int i=head[u]; i!=-1; i=edge[i].next)

        {

            if(edge[i].cap && dis[edge[i].v] < (dis[u]+edge[i].w))//求最长路

            {

                dis[edge[i].v] = dis[u]+edge[i].w;

                pre[edge[i].v] = i;

                if(!vis[edge[i].v])

                {

                    vis[edge[i].v]=1;

                    q.push(edge[i].v);

                }

            }

        }

    }

    if(dis[t] != -1)//------------------忘改了。

。

return 1;

    else

        return 0;

}

int MinCostMaxFlow(int s,int t)

{

    int flow=0,cost=0;

    while(spfa(s,t))

    {

        int df = inf;

        for(int i = pre[t]; i!=-1; i=pre[edge[i].u])

        {

            if(edge[i].cap<df)

                df = edge[i].cap;

        }

        flow += df;

        for(int i=pre[t]; i!=-1; i=pre[edge[i].u])

        {

            edge[i].cap -= df;

            edge[i^1].cap += df;

        }

        //printf("df = %d\n",df);

        cost += dis[t] * df;

    }

    return cost;

}

void initt()

{

    cnt=0;

    memset(head,-1,sizeof(head));

}

int ma;

int main()

{

    int s,t,k;

    while(~scanf("%d",&n))

    {

        initt();

        s = 0;

        t=2*n*n+1;

        add(s,1,0,2);

        int num = n*n;

        FOR(i,1,n+1)

        {

            FOR(j,1,n+1)

            {

                scanf("%d",&ma);

                add((i-1)*n+j,(i-1)*n+j+num,ma,1);

                add((i-1)*n+j,(i-1)*n+j+num,0,1);//本点与拆点连线，费用0。容量为无穷

                if(i<=n-1)//向下建图

                {

                    add((i-1)*n+j+num,i*n+j,0,1);

                }

                if(j<=n-1)//向右建图

                {

                     add((i-1)*n+j+num,(i-1)*n+j+1,0,1);

                }

            }

        }

        add(t-1,t,0,2);

        int ans =  MinCostMaxFlow(s,t);

        printf("%d\n",ans);

    }

    return 0;

}