[题解] Codeforces 429-B

B. Working out
time limit per test:2 seconds
memory limit per test:256 megabytes
input:standard input
output:standard output

Contents

1 Description
2 Input
3 Output
4 Sample Input
5 sample Output
6 Note
- 6.1 代码

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 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.

Sample Input

3 3
100 100 100
100 1 100
100 100 100

sample 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].

这道题我们通过构建4个dp数组来解决，我们假设两人在[i][j]处相遇

dp1[n][m] 表示从[1][1] 到 [i][j] 的最大收益
dp2[n][m] 表示从[i][j] 到 [n][m] 的最大收益
dp3[n][m] 表示从[n][1] 到 [i][j] 的最大收益
dp4[n][m] 表示从[i][j] 到 [1][m] 的最大收益

然后我们枚举可能相遇的点[i][j]，因为只能相遇一次，所以相遇时可能有以下两种情况

情况1:Iahub向下走，Iahubina向右走（如果向上或向下都有多个重复，不能向左走）
情况2:Iahub向游走，Iahubina向上走（同理）

最后输出答案就可以了!

代码

int main(){
    scanf("%d%d",&n,&m);

    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            scanf("%d",&data[i][j]);

    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            dp1[i][j] = max(dp1[i][j-1],dp1[i-1][j]) + data[i][j];

    for(int i=n;i>=1;i--)
        for(int j=m;j>=1;j--)
            dp2[i][j] = max(dp2[i][j+1],dp2[i+1][j]) + data[i][j];

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

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

    int ans = 0;

    for(int i=2;i

BLOG