C++ :

//方法一：四维
#include<bits/stdc++.h>
using namespace std;

int dp[12][12][12][12],a[12][12];
int main()
{
	int n,x,y,z;
	cin>>n;
	while(cin>>x>>y>>z&&x&&y&&z){
		a[x][y]=z;
	}
	for(int x1=1; x1<=n; x1++){
		for(int y1=1; y1<=n; y1++){
			for(int x2=1; x2<=n; x2++){
				for(int y2=1; y2<=n; y2++){
					dp[x1][y1][x2][y2]=max(dp[x1-1][y1][x2-1][y2],
					                   max(dp[x1-1][y1][x2][y2-1],
									   max(dp[x1][y1-1][x2-1][y2],
                                       dp[x1][y1-1][x2][y2-1])))
                                       +a[x1][y1]+a[x2][y2];
					if(x1==x2&&y1==y2){
						dp[x1][y1][x2][y2]-=a[x1][y1];//多加了1次 
					} 
				}
			}
		}
	}
	cout<<dp[n][n][n][n]<<endl; 
	return 0;
}

/*分析
状态：dp[x1][y1][x2][y2]（一个走到x1,y1,一个走到x2,y2的最大路径和）
转移方程：
dp[x1][y1][x2][y2]=max(dp[x1-1][y1][x2-1][y2],
				   max(dp[x1-1][y1][x2][y2-1],
		           max(dp[x1][y1-1][x2-1][y2],
                       dp[x1][y1-1][x2][y2-1])))
                        +a[x1][y1]+a[x2][y2]; 
边界值：无
计算顺序 dp[1..n][1...n]从小到大
答案：dp[n][n][n][n]
*/

Pascal :

program fgqs;
var a:array[0..100,0..100] of longint;
    f:array[0..100,0..100,0..100] of longint;
    i,j,n,x,y,t:longint;
function max(b,c,d,e:longint):longint;
begin
  if c>b then b:=c;
  if d>b then b:=d;
  if e>b then b:=e;
  exit(b);
end;
begin
 { assign(input,'fgqs.in');
  assign(output,'fgqs.out');
  reset(input);
  rewrite(output);}
  readln(n);
  readln(x,y,t);
  while (x<>0) and (y<>0) and (t<>0) do
  begin
  a[x,y]:=t;
    readln(x,y,t);
  end;
  f[1,1,1]:=a[1,1];
  for i:=2 to n+n-1 do
    for x:=1 to n do
      for y:=1 to n do
        if (i+1-x>0) and (i+1-y>0) then
        begin
          f[i,x,y]:=max(f[i-1,x-1,y],f[i-1,x-1,y-1],f[i-1,x,y-1],f[i-1,x,y]);
          if x<>y then
          f[i,x,y]:=f[i,x,y]+a[x,i+1-x]+a[y,i+1-y];
          if x=y then
          f[i,x,y]:=f[i,x,y]+a[x,i+1-x];
        end;
  writeln(f[n+n-1,n,n]);
  {close(input);
  close(output);}
end.