CodeForces 429B

 Working out

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

 

Description

Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in thei-th line and the j-th column.

Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workouta[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].

There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.

If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.

Input

The first line of the input contains two integers n and m (3 ≤ n, m ≤ 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 ≤ a[i][j] ≤ 10⁵).

Output

The output contains a single number — the maximum total gain possible.

Sample Input

Input

3 3
100 100 100
100 1 100
100 100 100

Output

800

Hint

Iahub will choose exercises a[1][1] → a[1][2] → a[2][2] → a[3][2] → a[3][3]. Iahubina will choose exercises a[3][1] → a[2][1] → a[2][2] → a[2][3] → a[1][3].

题目意思：给一个n*m的网格，一个人从左上走到右下（只能往右或往下），另一个人从左下走到右上（只能往右或往上），每个格子都有一定数值，经过就可以获得该数值，两人在网格中只能相遇一次，相遇点数值两人都不能获得，求两人获得数值和的最大值。 

解题思路：先预处理出从四个顶点出发到任一点的最大值，用dp即可。然后 枚举相遇点，相遇点不可能在边缘，因为那样交点就肯定不止一个，对于每个交点有两种方式走，求出两种方式最大值就可。

//2016.8.30
#include<iostream>
#include<cstdio>

using namespace std;

const int maxn = 1005;
int n, m, a[maxn][maxn];
int dp1[maxn][maxn],dp2[maxn][maxn],dp3[maxn][maxn],dp4[maxn][maxn];
//dp1[i][j] := 从 (1, 1) 到 (i, j) 的最大分数 
//dp2[i][j] := 从 (i, j) 到 (n, m) 的最大分数 
//dp3[i][j] := 从 (n, 1) 到 (i, j) 的最大分数 
//dp4[i][j] := 从 (i, j) 到 (1, m) 的最大分数
int main()
{
    while(cin>>n>>m)
    {
        for(int i = 1; i <= n; i++)
          for(int j = 1; j <= m; j++)
            scanf("%d", &a[i][j]);
    for(int i = 1; i <= n; i++)
      for(int j = 1; j <= m; j++)
        dp1[i][j] = max(dp1[i-1][j], dp1[i][j-1])+a[i][j];
    
    for(int i = n; i >= 1; i--)
       for(int j = m; j >= 1; j--)
        dp2[i][j] = max(dp2[i+1][j], dp2[i][j+1])+a[i][j];

    for(int i = n; i >= 1; i--)
        for(int j = 1; j <= m; j++)
        dp3[i][j] = a[i][j]+max(dp3[i][j-1], dp3[i+1][j]);

    for(int i = 1; i <= n; i++)
        for(int j = m; j >= 1; j--)
        dp4[i][j] = a[i][j]+max(dp4[i][j+1], dp4[i-1][j]);

    int ans = 0;
    for(int i = 2; i < n; i++)
      for(int j  = 2; j < m; j++)
      {
          ans = max(ans, dp1[i-1][j]+dp2[i+1][j]+dp3[i][j-1]+dp4[i][j+1]);
          ans = max(ans, dp1[i][j-1]+dp2[i][j+1]+dp3[i+1][j]+dp4[i-1][j]);
      }
    cout<<ans<<endl;
    }
    return 0;
}