Codeforces 214E Relay Race【Dp】

E. Relay Race

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Furik and Rubik take part in a relay race. The race will be set up on a large square with the side of n meters. The given square is split into n × n cells (represented as unit squares), each cell has some number.

At the beginning of the race Furik stands in a cell with coordinates (1, 1), and Rubik stands in a cell with coordinates (n, n). Right after the start Furik runs towards Rubik, besides, if Furik stands at a cell with coordinates (i, j), then he can move to cell (i + 1, j) or (i, j + 1). After Furik reaches Rubik, Rubik starts running from cell with coordinates (n, n) to cell with coordinates (1, 1). If Rubik stands in cell (i, j), then he can move to cell (i - 1, j) or (i, j - 1). Neither Furik, nor Rubik are allowed to go beyond the boundaries of the field; if a player goes beyond the boundaries, he will be disqualified.

To win the race, Furik and Rubik must earn as many points as possible. The number of points is the sum of numbers from the cells Furik and Rubik visited. Each cell counts only once in the sum.

Print the maximum number of points Furik and Rubik can earn on the relay race.

Input

The first line contains a single integer (1 ≤ n ≤ 300). The next n lines contain n integers each: the j-th number on the i-th line ai, j( - 1000 ≤ ai, j ≤ 1000) is the number written in the cell with coordinates (i, j).

Output

On a single line print a single number — the answer to the problem.

Examples

input

15

output

5

input

211 1416 12

output

53

input

325 16 2512 18 1911 13 8

output

136

Note

Comments to the second sample: The profitable path for Furik is: (1, 1), (1, 2), (2, 2), and for Rubik: (2, 2), (2, 1), (1, 1).

Comments to the third sample: The optimal path for Furik is: (1, 1), (1, 2), (1, 3), (2, 3), (3, 3), and for Rubik: (3, 3), (3, 2), (2, 2), (2, 1), (1, 1). The figure to the sample:

Furik's path is marked with yellow, and Rubik's path is marked with pink.

题目大意：

从左上角只能向右向下走，有两个人一起走，求走过的格子的价值和，一个格子走过多次只能积累一次和。

问最大能够积累多大的和。

思路：

因为存在负权值，所以这个问题我们不能考虑用费用流去做。

我们考虑Dp。最简单最暴力的方法就是去设定Dp【i】【j】【x】【y】表示第一个人走到了（i，j）这个点，另一个人走到了（x，y）这个点能够获得的最优价值和。

但是考虑n比较大，O（n^4）去做不仅空间炸了，时间也炸了。

那么考虑设定Dp【i】【x】【step】表示第一个人走到了第i行，第二个人走到了第x行，一共走了step步的最优价值和，那么有：

当前step步，第一个人位子为：i，2+step-i，第二个人的位子为：x，2+step-x；

那么对于两个人，一共四种走法，转移一下有：

过程维护一下即可。

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int a[350][350];int dp[305][305][305*2+5];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<=n;i++)        {            for(int j=0;j<=n;j++)            {                for(int k=0;k<=2*n;k++)                {                    dp[i][j][k]=-0x3f3f3f3f;                }            }        }        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                scanf("%d",&a[i][j]);            }        }        dp[1][1][0]=a[1][1];        for(int i=1;i<=n;i++)        {            for(int x=1;x<=n;x++)            {                for(int step=1;step<=2*n-1;step++)                {                    int j=2+step-i;                    int y=2+step-x;                    if(i==1&&j==1)continue;                    if(x==1&&y==1)continue;                    if(j>=1&&j<=n&&y>=1&&y<=n)                    {                        dp[i][x][step]=max(dp[i][x][step],dp[i-1][x-1][step-1]+a[i][j]+a[x][y]);                        dp[i][x][step]=max(dp[i][x][step],dp[i-1][x][step-1]+a[i][j]+a[x][y]);                        dp[i][x][step]=max(dp[i][x][step],dp[i][x-1][step-1]+a[i][j]+a[x][y]);                        dp[i][x][step]=max(dp[i][x][step],dp[i][x][step-1]+a[i][j]+a[x][y]);                        if(i==x&&j==y)dp[i][x][step]-=a[x][y];                    }                }            }        }        printf("%d\n",dp[n][n][2*n-2]);    }}