Codeforces

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问题

题目描述

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.

Example 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].

题解

分析

  • 题目大意是两个人分别从一个角出发走向对角,求两个人只相遇一个点得到的最大权值。
  • 相遇点一定不在边缘,所以,我们要枚举所有不在边缘的一个点中的最大值。枚举之前,先dp,dp4次,分别求从角落到每个点的距离,从(1,1)到(n,m)和逆向的(n,m)到(1,1)。这样我们就可以算出某个点最大值。看代码。

代码

#include<iostream>#include<algorithm>#include<cstdio>using namespace std;const int maxn=1e3+5;LL a[maxn][maxn];LL dp1[maxn][maxn];LL dp2[maxn][maxn];LL dp3[maxn][maxn];LL dp4[maxn][maxn];int main(){    int n,m;    while(~scanf("%d%d",&n,&m)){        memset(dp1,0,sizeof(dp1));        memset(dp2,0,sizeof(dp2));        memset(dp3,0,sizeof(dp3));        memset(dp4,0,sizeof(dp4));        for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++)            scanf("%d",&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]);        LL ans=0;        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("%lld\n",ans);    }    return 0;}