题目地址：https://vjudge.net/contest/237921#problem/B
Working out(动态规划)

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 the i-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 workout a[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.

Examples
Input
3 3
100 100 100
100 1 100
100 100 100
Output
800

Note
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].

题目大意：
一个n*m的矩阵，一个人从左上角走到右下角，一个人从左下角走到右上角。每个人走到一个格子会获得格子对应的数字。两个人必须且只能相遇一次。当相遇时都不能获得该格子对应的数字。求两人累计所获得的数字的最大和。

解题思路：
先求出
dp1[i][j]=max(dp1[i-1][j],dp1[i][j-1])+a[i][j] 表示从1，1到i，j的最大值
dp2[i][j]=max(dp2[i+1][j],dp2[i][j+1])+a[i][j] 表示从 i， j到n，m的最大值
dp3[i][j]=max(dp3[i+1][j],dp3[i][j-1])+a[i][j] 表示从n，1到i，j的最大值
dp4[i][j]=max(dp4[i-1][j],dp4[i][j+1])+a[i][j] 表示从i，j到1，m的最大值

由于只能相遇一次故，到相遇点时只有两种情况，一种是第一个人横着走过去，第二个人竖着走过去，然后反过来。
代码如下：

#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef long long ll;
const int maxn=1000+5;
int a[maxn][maxn];
int dp1[maxn][maxn],dp2[maxn][maxn],dp3[maxn][maxn],dp4[maxn][maxn]; 
int n,m;
int main(){
    ios::sync_with_stdio(false);
    while(cin>>n>>m)
    {
        memset(dp1,0,sizeof(dp1));
        memset(dp3,0,sizeof(dp3));
        memset(dp2,0,sizeof(dp2));
        memset(dp4,0,sizeof(dp4));
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                cin>>a[i][j];

        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                dp1[i][j]=a[i][j]+max(dp1[i-1][j],dp1[i][j-1]);
            }
        }
        for(int i=n;i>=1;i--){
            for(int j=m;j>=1;j--){
                dp2[i][j]=a[i][j]+max(dp2[i+1][j],dp2[i][j+1]);
            }
        }
        for(int i=n;i>=1;i--){
            for(int j=1;j<=m;j++){
                dp3[i][j]=a[i][j]+max(dp3[i+1][j],dp3[i][j-1]);
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=m;j>=1;j--){
                dp4[i][j]=a[i][j]+max(dp4[i-1][j],dp4[i][j+1]);
            }
        }
        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;
}