블루 브리지 컵 - 알고리즘 트레이닝 격자 추출수(다중 스레드 dp)

#include 
#include 
#include 
#include 
using namespace std;

const int maxn = 12;
int   dp[maxn][maxn][maxn][maxn], g[maxn][maxn];
int   n;

void dpSolve()
{
    /*for(int k = 2; k < n + n; ++k)
    {
        for(int x1 = 1; x1 <= n; ++x1){
            for(int x2 = 1; x2 <= n; ++x2){

                int y1 = k - x1, y2 = k - x2;
                if(y1 < 1 || y1 > n || y2 < 1 || y2 > n) continue;
                dp[k][x1][x2] = max(max(dp[k - 1][x1][x2], dp[k - 1][x1 - 1][x2 - 1]),
                                    max(dp[k - 1][x1 - 1][x2], dp[k - 1][x1][x2 - 1]));
                if(y1 == y2) dp[k][x1][x2] += g[x1][y1];
                else dp[k][x1][x2] += g[x1][y1] + g[x2][y2];
            }
        }
    }
    cout << dp[n + n - 1][n][n - 1] << endl;*/
    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(max(dp[x1-1][y1][x2-1][y2], dp[x1-1][y1][x2][y2-1]),
                                             max(dp[x1][y1-1][x2-1][y2],dp[x1][y1-1][x2][y2-1]));
                    if(x1 == x2 && y1 == y2) dp[x1][y1][x2][y2] += g[x1][y2];
                    else dp[x1][y1][x2][y2] += g[x1][y1] + g[x2][y2];
                }
    }
    cout << dp[n][n][n][n] << endl;
}

int main()
{
    #ifdef LOCAL
    freopen("data.in", "r", stdin);
    #endif // LOCAL
    cin >> n;
    int a, b, c;
    while(cin >> a >> b >> c)
    {
        if(!a && !b && !c) break;
        g[a][b] = c;
    }
    dpSolve();
    return 0;
}