CF 429B (三维dp)

最新推荐文章于 2022-03-15 22:53:44 发布

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本文介绍了一个关于寻找两个人在矩阵中行走以获得最大总价值的问题。通过动态规划的方法，计算了从矩阵的不同角落出发到达任意一点的最大累积价值，最终确定了两人相遇点能够达到的最大累积价值。

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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.

Examples

Input

3 3
100 100 100
100 1 100
100 100 100

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].

题目大意：

有一个矩阵，一个人从左上往右下走，一个人从左下往右上走，中间有一次相遇该点的值不算，求最后两人走过路线值之和的最大值（两人重复走过的点只计算一次）

解题思路：

用三维一个三维矩阵记录矩阵上的每个点分别到矩阵上四个角最大的值，假设某点是相遇点，该路线的值就是该点到四个角上的最大值之和。因此遍历矩阵上的每个点，求出最大值即可。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
long long a[1005][1005],dp[5][1005][1005];
int main()
{
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			scanf("%lld",&a[i][j]);
	memset(dp,0,sizeof(dp));
	for(int i=1;i<=n;i++)//dp各点到四个角的最大距离
	{
		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 j=m;j>=1;j--)
			dp[3][i][j]=max(dp[3][i-1][j],dp[3][i][j+1])+a[i][j];
	}
	for(int i=n;i>=1;i--)
	{
		for(int j=1;j<m;j++)
			dp[2][i][j]=max(dp[2][i+1][j],dp[2][i][j-1])+a[i][j];
		for(int j=m;j>=1;j--)
			dp[4][i][j]=max(dp[4][i+1][j],dp[4][i][j+1])+a[i][j];
	}
	long long sum=0;
	for(int i=2;i<n;i++)//因为走重复的点肯定亏损，所有只有两种走法并且只走内圈
		for(int j=2;j<m;j++)
		{
			sum=max(sum,dp[1][i-1][j]+dp[4][i+1][j]+dp[2][i][j-1]+dp[3][i][j+1]);
			sum=max(sum,dp[1][i][j-1]+dp[4][i][j+1]+dp[2][i+1][j]+dp[3][i-1][j]);
		}
	printf("%lld\n",sum);
    return 0;
}