Codeforces 429B Working out (dp 数字三角形变形)

Working out

time limit per test：2 seconds

memory limit per test：256 megabytes

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 matrixa with n lines andm 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 workouta[n][m]. After finishing workouta[i][j], he can go to workouta[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workouta[n][1] and she needs to finish with workouta[1][m]. After finishing workout from cella[i][j], she goes to eithera[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 integersn and m (3 ≤ n, m ≤ 1000). Each of the nextn lines contains m integers:j-th number from i-th line denotes elementa[i][j] (0 ≤ a[i][j] ≤ 10⁵).

Output

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

Examples

Input

3 3100 100 100100 1 100100 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 exercisesa[3][1] → a[2][1] → a[2][2] → a[2][3] → a[1][3].

题目链接：http://codeforces.com/problemset/problem/429/B

题目大意：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

题目分析：四个角求四次，对于两个人枚举两种情况(A从上往下B从左往右，A从左往右B从下往上)

#include <cstdio>#include <algorithm>#define ll long longusing namespace std;int const MAX = 1e3 + 5;ll a[MAX][MAX], dp1[MAX][MAX], dp2[MAX][MAX], dp3[MAX][MAX], dp4[MAX][MAX], ans;int main(){    int n, m;    scanf("%d %d", &n, &m);    for(int i = 1; i <= n; i++)        for(int j = 1; j <= m; j++)            scanf("%I64d", &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]);        }    }    printf("%I64d\n", ans);}