把問題抽象成圖論問題，數學模型是求從S到T的兩條不相交的路徑，使得路徑上點的權值之和最大。

費用流建模，首先拆點，把頂點i拆成i.a和i.b，i.a 與i.b之間連接一條費用爲“好心值”，容量爲1的有向邊，特殊地，左上角和右下角兩個節點拆分後點內邊容量設爲2，因爲我們要找兩條不相交路徑。i右邊和下邊的節點j，連接一條(i.b,j.a)費用爲0，容量爲1的有向邊。

求最大費用最大流即可，費用流值就是要求的結果。比動態規劃運行得快，空間佔用少。

/* 
 * Problem: NOIP2008 message
 * Author: Guo Jiabao
 * Time: 2009.6.24 11:52
 * State: Solved
 * Memo: 最小費用最大流
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
using namespace std;
const int MAXL=51,MAXN=MAXL*MAXL*2,MAXM=MAXN*3,INF=~0U>>1;
struct Queue
{
    int Q[MAXN],head,tail,size;
    bool inq[MAXN];
    Queue()
    {
        memset(inq,0,sizeof(inq));
        head=size =0;
        tail=-1;
    }
    void ins(int p)
    {
        inq[p]=true;
        if (++tail >= MAXN) tail = 0;
        Q[tail] = p;
        size ++;
    }
    int pop()
    {
        int p=Q[head];
        if (++head >= MAXN) head = 0;
        inq[p]=false;
        size --;
        return p;
    }
}Q;
struct edge
{
    edge *next,*op;
    int t,c,v;
}ES[MAXM],*V[MAXN],*fe[MAXN];
int N,M,EC,S,T,CostFlow;
int dist[MAXN],ft[MAXN];
inline void addedge(int a,int b,int v)
{
    ES[++EC].next = V[a]; V[a]=ES+EC; V[a]->t = b; V[a]->c=1; V[a]->v = v;
    ES[++EC].next = V[b]; V[b]=ES+EC; V[b]->t = a; V[b]->c=0; V[b]->v = -v;
    V[a]->op = V[b]; V[b]->op = V[a];
}
void init()
{
    int i,j,a,b,c;
    freopen("message.in","r",stdin);
    freopen("message.out","w",stdout);
    scanf("%d%d",&N,&M);
    for (i=1;i<=N;i++)
    {
        for (j=1;j<=M;j++)
        {
            a=(i-1)*M+j;b=a+a;a=b-1;
            scanf("%d",&c);
            addedge(a,b,c);
        }
    }
    ES[EC-1].c=ES[1].c=2;
    for (i=1;i<=N;i++)
    {
        for (j=1;j<=M;j++)
        {
            a=((i-1)*M+j)*2;
            if (j+1<=M)
            {
                b = ((i-1)*M+j+1)*2 - 1;
                addedge(a,b,0);
            }
            if (i+1<=N)
            {
                b = (i*M+j)*2 - 1;
                addedge(a,b,0);
            }
        }
    }    
    S=1; T = N * M * 2;
}
bool spfa()
{
    int i,j;
    for (i=S;i<=T;i++)
        dist[i]=-INF;
    dist[S]=0;
    Q.ins(S);
    while (Q.size)
    {
        i= Q.pop();
        for (edge *e=V[i];e;e=e->next)
        {
            j=e->t;
            if (e->c && dist[i] + e->v > dist[j])
            {
                dist[j] = dist[i] + e->v;
                ft[j] = i;
                fe[j] = e;
                if (!Q.inq[j])
                    Q.ins(j);
            }
        }
    }
    return dist[T]!=-INF;
}
void aug()
{
    int i,delta=INF;
    for (i=T;i!=S;i=ft[i])
        if (fe[i]->c < delta)
            delta = fe[i]->c;
    for (i=T;i!=S;i=ft[i])
    {
        fe[i]->c -= delta;
        fe[i]->op->c +=delta;
        CostFlow += fe[i]->v * delta;
    }
}
void solve()
{
    while (spfa())
        aug();
}
int main()
{
    init();
    solve();
    printf("%d\n",CostFlow);
    return 0;
}