CodeForces 429B Working out(dp,四个角递推)

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
[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
[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.

Example

Input

3 3

100 100 100

100 1 100

100 100 100

Output

800

题意：两个人，一个从左上到右下，一个从左下到右上，要求只相遇一个点，相遇的点权值不算，求最大总权值

思路：DP枚举出每一个点到四个角的最大值，
dp1[i][j] 从（1,1）到（i,j）的最大值
dp2[i][j] 从（n,1）到（i,j）的最大值
dp3[i][j] 从（1,m）到（i,j）的最大值
dp1[i][j] 从（m,n）到（i,j）的最大值

对于两个人枚举两种情况(因为有一个限制条件2者只可以相遇一次)

1)A从上往下B从左往右

2)A从左往右B从下往上，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdio>
using namespace std;
typedef long long ll;
#define N 1005
ll dp1

,dp2

,dp3

,dp4

;
ll a

;
int n,m;
int main()
{
scanf("%d%d",&n,&m);
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
scanf("%lld",&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 = 1; j <= m; j++)
dp2[i][j] = a[i][j] + max(dp2[i + 1][j], dp2[i][j - 1]);
for(int i = 1; i <= n; i++)
for(int j = m; j >= 1; j--)
dp3[i][j] = a[i][j] + max(dp3[i - 1][j], dp3[i][j + 1]);
for(int i = n; i >= 1; i--)
for(int j = m; j >= 1; j--)
dp4[i][j] = a[i][j] + max(dp4[i + 1][j], dp4[i][j + 1]);
for(int i = 2; i < n; i++)
{
for(int j = 2; j < m; j++)
{
ans = max(ans, dp1[i - 1][j] + dp4[i + 1][j] + dp2[i][j - 1] + dp3[i][j + 1]);
ans = max(ans, dp1[i][j - 1] + dp4[i][j + 1] + dp2[i + 1][j] + dp3[i - 1][j]);
}
}
}