作者：Accagain

原題

B. Working out

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output 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].

題意

給n*m的矩陣，每個格子有個數，A從(1,1)出發只能向下或右走，終點為(n,m)，B從(n,1)出發只能向上或右走，終點為(1,m)。兩個人的速度不一樣，走到的格子可以獲的該格子的數，兩人相遇的格子上的數兩個人都不能拿。求A和B能拿到的數的總和的最大值。

涉及知識及演算法

動態規劃 先計算出每個格子到四個角落格子的路徑最大數值，然後列舉兩個人相遇的交點格子，從交點格子到四個角落的路徑之和即為題目要求的總和。不過要注意的是，有兩種方式匯於交點，畫出來大概像佛教“萬”字卍和法西斯(納粹)符號“卐“，求總和的最大值即可。需要注意邊界情況。

程式碼

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 1100
ll dp[5][Maxn][Maxn];
int n,m;

ll save[Maxn][Maxn];


int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   while(~scanf("%d%d",&n,&m))
   {
       memset(save,0,sizeof(save));

       for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%d",&save[i][j]);

       memset(dp,0,sizeof(dp));
       for(int i=1;i<=n;i++)
       {
           //從左上角
           for(int j=1;j<=m;j++)
               dp[1][i][j]=max(dp[1][i-1][j],dp[1][i][j-1])+save[i][j];
               //printf("i:%d j:%d %I64d\n",i,j,dp[1][i][j]);
           //從右上角
           for(int j=m;j>=1;j--)
                dp[3][i][j]=max(dp[3][i-1][j],dp[3][i][j+1])+save[i][j];
       }

       for(int i=n;i>=1;i--)
       {
           //從左下角
           for(int j=1;j<=m;j++)
                dp[2][i][j]=max(dp[2][i+1][j],dp[2][i][j-1])+save[i][j];
           //從右下角
           for(int j=m;j>=1;j--)
                dp[4][i][j]=max(dp[4][i+1][j],dp[4][i][j+1])+save[i][j];
       }
       ll ans=0;
       //把四個邊界都置為無效情況
       for(int i=0;i<=n+1;i++)  
            for(int k=1;k<=4;k++)
                dp[k][i][0]=dp[k][i][m+1]=-INF;
       for(int j=0;j<=m+1;j++)
            for(int k=1;k<=4;k++)
                dp[k][0][j]=dp[k][n+1][j]=-INF;

       for(int i=1;i<=n;i++)
       {
           for(int j=1;j<=m;j++)
           {    
               //只用考慮兩種情況 兩個人只能交叉一個格子，多了的話不合算
               //卐形
               ll temp=dp[1][i][j-1]+dp[4][i][j+1]+dp[2][i+1][j]+dp[3][i-1][j];
               ans=max(ans,temp);
               //卍形
               temp=dp[1][i-1][j]+dp[4][i+1][j]+dp[2][i][j-1]+dp[3][i][j+1];
               ans=max(ans,temp);
               //printf("i:%d j:%d %I64d\n",i,j,ans);
           }
       }
       printf("%I64d\n",ans);
   }
   return 0;
}