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.

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).

The output contains a single number — the maximum total gain possible.

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].

题意就是两个人，一个人从左上角走到右下角，只能往下往右走，一个人从左下角走到右上角，只能往上往右走，中间俩人要经过一个共同的地方，这个地方的值不加，两人走过的其他地方加起来，两人除了一定要共同经过一个地方，其他地方只能有一个人经过一次。

思路：

先dp四次，分别从左上角走到右下角，左下角走到右上角，右上角走到左下角，右下角走到左上角。

然后考虑两人相遇的情况，只有两种满足，一种是右上，一种是下右，相遇之后也是这种状态。

然后枚举点，边界处是不满足的，去掉。

代码:

 //Codeforces 429 B. Working out-dp

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 const int maxn=1e3+;

 const int maxnm=1e5+;

 ll dp1[maxn][maxn],dp2[maxn][maxn],dp3[maxn][maxn],dp4[maxn][maxn];

 int main()

 {

     int n,m;

     scanf("%d%d",&n,&m);

     int a[n+][m+];

     for(int i=;i<=n;i++){

         for(int j=;j<=m;j++)

             scanf("%d",&a[i][j]);

     }

     for(int i=;i<=n;i++){//(1,1)->(n,m)

         for(int j=;j<=m;j++){

             dp1[i][j]=max(dp1[i-][j],dp1[i][j-])+a[i][j];

         }

     }

     for(int i=n;i>;i--){//(n,1)->(1,m)

         for(int j=;j<=m;j++){

             dp2[i][j]=max(dp2[i][j-],dp2[i+][j])+a[i][j];

         }

     }

     for(int i=n;i>;i--){//(n,m)->(1,1)

         for(int j=m;j>;j--){

             dp3[i][j]=max(dp3[i][j+],dp3[i+][j])+a[i][j];

         }

     }

     for(int i=;i<=n;i++){//(1,m)->(n,1)

         for(int j=m;j>;j--){

             dp4[i][j]=max(dp4[i][j+],dp4[i-][j])+a[i][j];

         }

     }

     ll ans=;

     int posx,posy,flag=;

     for(int i=;i<n;i++){//枚举点，然后两种情况

         for(int j=;j<m;j++){//去掉边界，边界不满足情况，就可以了。

             ll cnt=dp1[i][j-]+dp3[i][j+]+dp2[i+][j]+dp4[i-][j];//右上

             ll ret=dp1[i-][j]+dp3[i+][j]+dp2[i][j-]+dp4[i][j+];//下右

             ans=max(ans,max(cnt,ret));

         }

     }

     printf("%lld\n",ans);

 }

 /*

 3 3

 3 1 2

 3 2 0

 2 3 2

 */

水博客时间结束。

最近不想写博客，不好玩。