cf429B dp递推

Posted 2020-08-14 a_clown_cz

Description

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 thei-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 workouta[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.

Sample Input

Input

3 3 100 100 100 100 1 100 100 100 100

Output

800

代码：注意一点，交点不能在边界，如果在边界不满足两人只相遇一次的条件。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

int mp[1005][1005];
int dp1[1005][1005],dp2[1005][1005],dp3[1005][1005],dp4[1005][1005];

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2)
    {
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
                scanf("%d",&mp[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++)//右下
            for(int j=1; j<=m; j++)
                dp1[i][j]=max(dp1[i-1][j],dp1[i][j-1])+mp[i][j];
        for(int i=n; i>=1; i--)//右上
            for(int j=1; j<=m; j++)
                dp2[i][j]=max(dp2[i+1][j],dp2[i][j-1])+mp[i][j];
        for(int i=1; i<=n; i++)//左下
            for(int j=m; j>=1; j--)
                dp3[i][j]=max(dp3[i-1][j],dp3[i][j+1])+mp[i][j];
        for(int i=n; i>=1; i--)//左上
            for(int j=m; j>=1; j--)
                dp4[i][j]=max(dp4[i+1][j],dp4[i][j+1])+mp[i][j];
        int ans=0;
        for(int i=2; i<n; i++)
            for(int j=2; j<m; j++)
            {
                ans=max(ans,dp1[i-1][j]+dp2[i][j-1]+dp3[i][j+1]+dp4[i+1][j]);
                ans=max(ans,dp1[i][j-1]+dp2[i+1][j]+dp3[i-1][j]+dp4[i][j+1]);
            }
        printf("%d\n",ans);
    }
    return 0;
}