1 #define _CRTDBG_MAP_ALLOC
  2 #include <stdlib.h>
  3 #include <crtdbg.h>
  4 void fnExit1(void)
  5 {
  6     _CrtDumpMemoryLeaks();
  7 }
  8
  9 #define _CRT_SECURE_NO_WARNINGS
 10 #define HOME
 11
 12 #include <iostream>
 13 #include <cstdlib>
 14 #include <cstdio>
 15 #include <cstddef>
 16 #include <iterator>
 17 #include <algorithm>
 18 #include <locale>
 19 #include <cmath>
 20 #include <vector>
 21 #include <cstring>
 22 using namespace std;
 23 const int INF = 0x3f3f3f3f;
 24 const int MaxN = 30;
 25 const int Max = 55;
 26
 27 int m, n;
 28 int arr[Max][Max];
 29 int dp[Max][Max][Max][Max];
 30 int step[2][2] = { {-1, 0}, {0, -1} }; // 向下，向左
 31
 32 bool Check(int x1, int y1, int x2, int y2)
 33 {
 34     // 防止越界
 35     if ((x1 <= 0) || (y1 <= 0) || (x2 <= 0) || (y2 <= 0))
 36     {
 37         return false;
 38     }
 39     // 两条路径不能相交
 40     if ((x1 == x2) && (y1 == y2))
 41     {
 42         return false;
 43     }
 44     return true;
 45 }
 46
 47 void Solve()
 48 {
 49     int x1, y1, x2, y2;
 50     for (int i = 1; i <= m; ++i)
 51     {
 52         for (int j = 1; j <= n; ++j)
 53         {
 54             for (int k = 1; k <= m; ++k)
 55             {
 56
 57                 // 两条路径并行走，则：(i+j)=(k+l)
 58                 if (i + j - k <= 0) break;
 59                 int l = i + j - k;
 60                 if (!Check(i, j, k, l)) continue;
 61                 int maxVal = 0;
 62                 for (int p = 0; p < 2; ++p)
 63                 {
 64                     for (int q = 0; q < 2; ++q)
 65                     {
 66                         x1 = i + step[p][0];
 67                         y1 = j + step[p][1];
 68                         x2 = k + step[q][0];
 69                         y2 = l + step[q][1];
 70                         if (Check(x1, y1, x2, y2))
 71                         {
 72                             maxVal = max(maxVal, dp[x1][y1][x2][y2]);
 73                         }
 74                     }
 75                 }
 76                 dp[i][j][k][l] = max(dp[i][j][k][l], maxVal + arr[i][j] + arr[k][l]);
 77             }
 78         }
 79     }
 80     // 两条路径要想到达点[m][n]，则两条路径的点必然选择：[m][n-1]，[m-1][n]
 81     int ans = max(dp[m][n - 1][m - 1][n], dp[m - 1][n][m][n-1]);
 82     //for (int i = 1; i <= m; ++i)
 83     //{
 84     //    for (int j = 1; j <= n; ++j)
 85     //    {
 86     //        for (int k = 1; k <= m; ++k)
 87     //        {
 88     //            for (int l = 1; l <= n; ++l)
 89     //            {
 90     //                //cout << "(" << i << ", " << j << ", " << k << ", " << l << ")" << " : " << dp[i][j][k][l] << endl;
 91     //                cout << dp[i][j][k][l] << "-->" << dp[k][l][i][j] << endl;
 92     //                 /*    两条路径出发点相同，没有区分两条路径，也就是小渊和小轩怎么分配这两条路径都可以，
 93     //                    所以dp[i][j][k][l]=dp[k][l][i][j]
 94     //                 */
 95     //            }
 96     //        }
 97     //    }
 98     //}
 99     cout << ans << endl;
100     //cout << dp[m][n - 1][m - 1][n] << " " << dp[m - 1][n][m][n - 1] << endl; // 两个值相同
101 }
102
103 int main()
104 {
105
106 #ifdef HOME
107     //freopen("in", "r", stdin);
108     //freopen("out", "w", stdout);
109     //atexit(fnExit1);
110 #endif
111
112     memset(dp, 0, sizeof(0));
113     cin >> m >> n;
114     for (int i = 1; i <= m; ++i)
115     {
116         for (int j = 1; j <= n; ++j)
117         {
118             cin >> arr[i][j];
119         }
120     }
121     Solve();
122
123
124 #ifdef HOME
125     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
126     _CrtDumpMemoryLeaks();
127     system("pause");
128 #endif
129     return 0;
130 }