CodeForces 429B Working out

题目链接：http://codeforces.com/problemset/problem/429/B

B. Working out

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Example

Input

3 3100 100 100100 1 100100 100 100

Output

800

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：A和B分别从1,1走向N,M，从N,1走向1，M。A只能向下或向右走，B只能向上或向右走。两人有且只有一次相遇，求两人所走路径的权值之和并使其最大，两人相遇的点上的权值不会被计算。

        分析题目可以得出，两人只能相遇一次，那么不可能在矩阵边缘处相遇，这样的话会导致多次相遇。对于相遇点我们只需枚举即可。根据AB移动方式我们可以得知两人只有两种方式，如图所

         

此外我们还需要DP出四个顶点到任一点的最大权值，四次DP即可。

然后对于任一相遇点（i,j）按以上两种方式更新最大值

cnt=max(cnt,dp1[i-1][j]+dp2[i+1][j]+dp4[i][j+1]+dp3[i][j-1]);

cnt=max(cnt,dp1[i][j-1]+dp2[i][j+1]+dp4[i-1][j]+dp3[i+1][j]);

最终AC代码如下：

#include <stdio.h>#include<string.h>#include<algorithm>using namespace std;int map[1005][1050]={0};int dp1[1005][1005]={0};int dp2[1005][1005]={0};int dp3[1005][1005]={0};int dp4[1005][1005]={0};int main(int argc, char *argv[]){int n,m;while(~scanf("%d %d",&n,&m)){memset(map,0,sizeof(map));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++)        for(int j=1;j<=m;j++)        {        dp1[i][j]=max(dp1[i-1][j],dp1[i][j-1])+map[i][j];        }        for(int i=n;i>=1;i--)        for(int j=m;j>=1;j--)        {        dp2[i][j]=max(dp2[i+1][j],dp2[i][j+1])+map[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])+map[i][j];        }        for(int i=n;i>=1;i--)        for(int j=1;j<=m;j++)        dp3[i][j]=max(dp3[i+1][j],dp3[i][j-1])+map[i][j];        int cnt=0;        for(int i=2;i<n;i++)        for(int j=2;j<m;j++)        {        cnt=max(cnt,dp1[i-1][j]+dp2[i+1][j]+dp4[i][j+1]+dp3[i][j-1]);        cnt=max(cnt,dp1[i][j-1]+dp2[i][j+1]+dp4[i-1][j]+dp3[i+1][j]);        }        printf("%d\n",cnt);}return 0;}

题目链接：http://codeforces.com/problemset/problem/429/B