CodeForces 429B Working out

Working out

Time Limit: 2000ms

Memory Limit: 262144KB

This problem will be judged on CodeForces. Original ID: 429B
64-bit integer IO format: %I64d      Java class name: (Any)

 

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 awith 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

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].

 

Source

Codeforces Round #245 (Div. 1)

 

解题：注意相遇的时候两人是不产生那个值的

我们可以枚举相遇点

其中左上角的人可以往右走来到相遇点后继续右走，左下角的人可以向上移动来到相遇点后继续上行

 

或者 左上角的人下移来到相遇点后继续下移，左下角的人右移来到相遇点 继续右移

 

至于最大值 简单dp

 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int dp[4][maxn][maxn],a[maxn][maxn],n,m;
int main() {
    while(~scanf("%d%d",&n,&m)) {
        memset(dp,0,sizeof dp);
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= m; ++j)
                scanf("%d",a[i] + j);
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= m; ++j)
                dp[0][i][j] = max(dp[0][i-1][j],dp[0][i][j-1]) + a[i][j];
        for(int i = n; i > 0; --i)
            for(int j = 1; j <= m; ++j)
                dp[1][i][j] = max(dp[1][i+1][j],dp[1][i][j-1]) + a[i][j];
        for(int i = 1; i <= n; ++i)
            for(int j = m; j > 0; --j)
                dp[2][i][j] = max(dp[2][i-1][j],dp[2][i][j+1]) + a[i][j];
        for(int i = n; i > 0; --i)
            for(int j = m; j > 0; --j)
                dp[3][i][j] = max(dp[3][i+1][j],dp[3][i][j+1]) + a[i][j];

        int ret = 0;
        for(int i = 2; i < n; ++i)
            for(int j = 2; j < m; ++j) {
                ret = max(ret,dp[0][i-1][j] + dp[3][i+1][j] + dp[1][i][j-1] + dp[2][i][j+1]);
                ret = max(ret,dp[0][i][j-1] + dp[3][i][j+1] + dp[1][i+1][j] + dp[2][i-1][j]);
            }
        printf("%d\n",ret);
    }
    return 0;
}