题意：
给一个矩形，从左上角走到右下角，并返回左上角（一个单元格只能走一次，左上角和右下角两个点除外）
并且从左上到右下只能往右和下两个方向。从右下返回左上只能走上和左两个方向！
分析：
拆点，最小费用最大流
。。
额。。。刘汝佳训练指南的最小费用最大流模板超时了。。。。。。。。。。。。。。。。。。
可能是因为边太少，点太多的缘故吧！还是数组实现的邻接表可靠啊！！！
// File Name: 3376.cpp
// Author: Zlbing
// Created Time: 2013年08月15日 星期四 13时24分37秒 #include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
/*
const int MAXN=720010; struct Edge{
int from,to,cap,flow,cost;
};
struct MCMF{
int n,m,s,t;
vector<Edge>edges;
vector<int> G[MAXN];
int inq[MAXN];
int d[MAXN];
int p[MAXN];
int a[MAXN];
void init(int n){
this->n=n;
for(int i=0;i<=n;i++)G[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int cap,int cost){
Edge e;
e.from=from,e.to=to,e.cap=cap,e.flow=0,e.cost=cost;
edges.push_back(e);
//edges.push_back((Edge){from,to,cap,0,cost});
e.from=to,e.to=from,e.cap=0,e.flow=0,e.cost=-cost;
//edges.push_back((Edge){to,from,0,0,-cost});
edges.push_back(e);
m=edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool BellmanFord(int s,int t,int& flow,int& cost){
for(int i=0;i<=n;i++)d[i]=INF;
CL(inq,0);
d[s]=0;inq[s]=1;p[s]=0;a[s]=INF; queue<int>Q;
Q.push(s);
while(!Q.empty()){
int u=Q.front();Q.pop();
inq[u]=0;
for(int i=0;i<(int)G[u].size();i++){
Edge& e=edges[G[u][i]];
if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
d[e.to]=d[u]+e.cost;
p[e.to]=G[u][i];
a[e.to]=min(a[u],e.cap-e.flow);
if(!inq[e.to]){
Q.push(e.to);
inq[e.to]=1;
}
}
}
}
if(d[t]==INF)return false;
flow+=a[t];
cost+=d[t]*a[t];
int u=t;
while(u!=s){
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
u=edges[p[u]].from;
}
return true;
}
int Mincost(int s,int t){
int flow=0,cost=0;
while(BellmanFord(s,t,flow,cost));
return cost;
}
};
MCMF solver;
*/
int sumFlow;
const int MAXN = ;
const int MAXM = ;
struct Edge
{
int u, v, cap, cost;
int next;
}edge[MAXM<<];
int NE;
int head[MAXN], dist[MAXN], pp[MAXN];
bool vis[MAXN];
void init()
{
NE = ;
memset(head, -, sizeof(head));
}
void addedge(int u, int v, int cap, int cost)
{
edge[NE].u = u; edge[NE].v = v; edge[NE].cap = cap; edge[NE].cost = cost;
edge[NE].next = head[u]; head[u] = NE++;
edge[NE].u = v; edge[NE].v = u; edge[NE].cap = ; edge[NE].cost = -cost;
edge[NE].next = head[v]; head[v] = NE++;
}
bool SPFA(int s, int t, int n)
{
int i, u, v;
queue <int> qu;
memset(vis,false,sizeof(vis));
memset(pp,-,sizeof(pp));
for(i = ; i <= n; i++) dist[i] = INF;
vis[s] = true; dist[s] = ;
qu.push(s);
while(!qu.empty())
{
u = qu.front(); qu.pop(); vis[u] = false;
for(i = head[u]; i != -; i = edge[i].next)
{
v = edge[i].v;
if(edge[i].cap && dist[v] > dist[u] + edge[i].cost)
{
dist[v] = dist[u] + edge[i].cost;
pp[v] = i;
if(!vis[v])
{
qu.push(v);
vis[v] = true;
}
}
}
}
if(dist[t] == INF) return false;
return true;
}
int MCMF(int s, int t, int n) // minCostMaxFlow
{
int flow = ; // 总流量
int i, minflow, mincost;
mincost = ;
while(SPFA(s, t, n))
{
minflow = INF + ;
for(i = pp[t]; i != -; i = pp[edge[i].u])
if(edge[i].cap < minflow)
minflow = edge[i].cap;
flow += minflow;
for(i = pp[t]; i != -; i = pp[edge[i].u])
{
edge[i].cap -= minflow;
edge[i^].cap += minflow;
}
mincost += dist[t] * minflow;
}
sumFlow = flow; // 题目需要流量，用于判断
return mincost;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
//solver.init(n*n*2+5);
init();
int a;
int s=;
int t=((n-)*n+n-)*+;
REP(i,,n-)
REP(j,,n-){
scanf("%d",&a);
//solver.AddEdge((i*n+j)*2,(i*n+j)*2+1,1,-a);
addedge((i*n+j)*,(i*n+j)*+,,-a);
if(i!=n-)
{
//solver.AddEdge((i*n+j)*2+1,((i+1)*n+j)*2,1,0);
addedge((i*n+j)*+,((i+)*n+j)*,,);
}
if(j!=n-)
{
//solver.AddEdge((i*n+j)*2+1,(i*n+j+1)*2,1,0);
addedge((i*n+j)*+,(i*n+j+)*,,);
}
}
addedge(s,s+,,);
addedge(t-,t,,);
int ans=-MCMF(s,t,t);
printf("%d\n",ans);
}
return ;
}