CF 429B (三维dp)

B. Working out

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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 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 column1. 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 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 integers n andm (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].

题目大意：

有一个矩阵，一个人从左上往右下走，一个人从左下往右上走，中间有一次相遇该点的值不算，求最后两人走过路线值之和的最大值（两人重复走过的点只计算一次）

解题思路：

用三维一个三维矩阵记录矩阵上的每个点分别到矩阵上四个角最大的值，假设某点是相遇点，该路线的值就是该点到四个角上的最大值之和。因此遍历矩阵上的每个点，求出最大值即可。

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string.h>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;long long a[1005][1005],dp[5][1005][1005];int main(){    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);int n,m;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)scanf("%lld",&a[i][j]);memset(dp,0,sizeof(dp));for(int i=1;i<=n;i++)//dp各点到四个角的最大距离{for(int j=1;j<=m;j++)dp[1][i][j]=max(dp[1][i-1][j],dp[1][i][j-1])+a[i][j];for(int j=m;j>=1;j--)dp[3][i][j]=max(dp[3][i-1][j],dp[3][i][j+1])+a[i][j];}for(int i=n;i>=1;i--){for(int j=1;j<m;j++)dp[2][i][j]=max(dp[2][i+1][j],dp[2][i][j-1])+a[i][j];for(int j=m;j>=1;j--)dp[4][i][j]=max(dp[4][i+1][j],dp[4][i][j+1])+a[i][j];}long long sum=0;for(int i=2;i<n;i++)//因为走重复的点肯定亏损，所有只有两种走法并且只走内圈for(int j=2;j<m;j++){sum=max(sum,dp[1][i-1][j]+dp[4][i+1][j]+dp[2][i][j-1]+dp[3][i][j+1]);sum=max(sum,dp[1][i][j-1]+dp[4][i][j+1]+dp[2][i+1][j]+dp[3][i-1][j]);}printf("%lld\n",sum);    return 0;}