Codeforces 429B Working out (动态规划)

A - Working out

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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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

Hint

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：

两个人在一个有 n×m 个房间的健身房内健身，每个房间都有一个锻炼可消耗的卡路里数。一个人从 a[1][1] 一路锻炼到 a[n][m] ，每次只能从 a[i][j] 移动到 a[i+1][j] 或 a[i][j+1]；另一个从 a[n][1] 一路锻炼到 a[1][m]，每次只能从 a[i][j] 移动到 a[i][j+1] 或 a[i-1][j] 。每在一个房间锻炼可以消耗对应的卡路里数，两个人必须在中间的某一个房间碰面且只能碰面一次，这个房间不计算卡路里数。求两人如何走，能使两人总计消耗卡路里数最大。

解题思路：

由于碰面点不确定且只有一次，可以设这点为（x，y），然后用DP分别求出从点（1，1），（n，m），（n，1），（1，m）出发到达点（x，y）消耗的最大卡路里数，最后再用一个二重循环找出满足题意的点（x，y）即可。

代码如下：

#include <iostream>

using namespace std;

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

int main()
{
    int n,m,i,j,ans;
    while(cin>>n>>m)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
             cin>>a[i][j];
        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=n;i>0;i--)
            for(int j=m;j>0;j--)
                dp2[i][j]=max(dp2[i+1][j],dp2[i][j+1])+a[i][j];
        for(int i=n;i>0;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=1;i<=n;i++)
            for(int j=m;j>0;j--)
                dp4[i][j]=max(dp4[i-1][j],dp4[i][j+1])+a[i][j];
        ans=-1;
        for(int i=2;i<n;i++)
           for(int j=2;j<m;j++)
           {
              ans=max(ans,dp1[i-1][j]+dp2[i+1][j]+dp3[i][j-1]+dp4[i][j+1]);
              ans=max(ans,dp1[i][j-1]+dp2[i][j+1]+dp3[i+1][j]+dp4[i-1][j]);
           }
        cout<<ans<<endl;
    }
    return 0;
}