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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 mcolumns. 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.
3 3
100 100 100 100 1 100 100 100 100
800
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].
一个人从左上走到右下，一个人从左下走到右上，两个人必须有一个点来相遇，而且相遇的点的值不能拿，问按照规则走，取得最大值的和为多少。递推dp。
可以枚举每一个点，假设每一个点都有可能相遇，然后找从(1,1),(1,m),(n,1)(n,m)四个点到（i，j）路径中得到的值的总和的最大值。
dp1[i][j] := 从 (1, 1) 到 (i, j) 的最大分数 dp[i][j]中存的就是左上到(i,j)点的最大值
dp2[i][j] := 从 (i, j) 到 (n, m) 的最大分数 dp[i][j]中存的就是(i,j)点到右下的最大值 dp3[i][j] := 从 (n, 1) 到 (i, j) 的最大分数 dp[i][j]中存的就是左下到(i,j)点的最大值 dp4[i][j] := 从 (i, j) 到 (1, m) 的最大分数 dp[i][j]中存的就是(i,j)点到右上的最大值#include#include #include #include #include using namespace std;const int maxn=1007;int dp1[maxn][maxn],dp2[maxn][maxn],dp3[maxn][maxn],dp4[maxn][maxn];int map[maxn][maxn];int main(){ int n,m; while(scanf("%d%d",&n,&m)!=EOF) { memset(dp1,0,sizeof(dp1)); memset(dp2,0,sizeof(dp2)); memset(dp3,0,sizeof(dp3)); memset(dp4,0,sizeof(dp4)); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%d",&map[i][j]); for(int i=1;i<=n;i++) //左上 for(int j=1;j<=m;j++) dp1[i][j]=max(dp1[i-1][j],dp1[i][j-1])+map[i][j]; //上一个位置是在左边或者是上面 for(int i=n;i>=1;i--) //左下 for(int j=1;j<=m;j++) dp2[i][j]=max(dp2[i+1][j],dp2[i][j-1])+map[i][j]; //上一个位置是在左边或者是下面 for(int i=1;i<=n;i++) //右上 for(int j=m;j>=1;j--) dp3[i][j]=max(dp3[i-1][j],dp3[i][j+1])+map[i][j]; //上一个位置是在右边或者是上面 for(int i=n;i>=1;i--) //右下 for(int j=m;j>=1;j--) dp4[i][j]=max(dp4[i+1][j],dp4[i][j+1])+map[i][j]; //上一个位置是在右边或者是下面 int ans=0; for(int i=2;i
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