code force 429B dp好题

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本文介绍了一个涉及矩阵中寻找两个不同起点终点的最优路径问题，并通过动态规划方法解决该问题，确保两条路径仅在一个格子中相遇，从而实现最大的权值总和。

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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

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].

题意：

给出一个矩形，让你求出从（1,1）到（n，m）的权值和加上（1，m）到（n，1）的权值和减去重合格子的权值

题解：

注意只能是重合一个格子，那么情况就只有下面两种了

那么我们可以先计算出点（i , j）到四个角的最大值

然后枚举这个矩形中的每一个点，计算最大值

因为要防止多个点重合，所以用下面的状态转移方程式

ans=max(ans,dp1[i][j-1]+dp2[i-1][j]+dp3[i+1][j]+dp4[i][j+1]);
ans=max(ans,dp1[i-1][j]+dp2[i][j+1]+dp3[i][j-1]+dp4[i+1][j]);

对于图一：要是重合的点在左上角或者右下角，那么就会产生错误的解（没有路从左下到右上）

对于图一：要是重合的点在左下角或者右上角，那么就会产生错误的解（没有路从左上到右下）

所以直接去除四个角的枚举范围即可

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define MAXN 1010
#define LL long long
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;
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
            scanf("%d",&a[i][j]);

        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++)
            dp1[i][j]=max(dp1[i-1][j],dp1[i][j-1])+a[i][j];
        for(int i=1;i<=n;i++)
            for(int j=m;j>=1;j--)
            dp2[i][j]=max(dp2[i-1][j],dp2[i][j+1])+a[i][j];
        for(int i=n;i>=1;i--)
            for(int j=1;j<=m;j++)
            dp3[i][j]=max(dp3[i+1][j],dp3[i][j-1])+a[i][j];
        for(int i=n;i>=1;i--)
            for(int j=m;j>=1;j--)
            dp4[i][j]=max(dp4[i+1][j],dp4[i][j+1])+a[i][j];

        LL ans=0;
        for(int i=2;i<n;i++){
            for(int j=2;j<m;j++){
                ans=max(ans,dp1[i][j-1]+dp2[i-1][j]+dp3[i+1][j]+dp4[i][j+1]);
                ans=max(ans,dp1[i-1][j]+dp2[i][j+1]+dp3[i][j-1]+dp4[i+1][j]);
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}