CodeForces 213C-Relay Race

题目

C. 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 ofn 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 nextn lines contain n integers each: thej-th number on the i-th line a_i, j( - 1000 ≤ a_i, 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.

题义

给出n*n的方格，每个格子上有1个数，有1个人从(1,1)走到(n,n)，只能往下或往右走，另一个从(n,n)到(1,1)，只能往上或往左走，当某个人经过格点时会得到格点的数，但一个格点子能被拿一次，问两个人能得到和的最大值。

解法

把问题转换为两个人都从(1,1)走到(n,n)，两个人同时走，设dp[i][j][k]表示两个人都走了i步，其中第一个人向右走了j步，第二个人向右走了k步

怎样就可以计算出两个人的当前坐标 x1=i-j+1,y1=j+1,x2=i-k+1,y2=k+1

然后往后递推，两个人分别可以往下或往右走，一共4种情况

答案是dp[2*n-2][n-1][n-1]

代码

#include<cstdio>#define INF 1000000000int dp[605][305][305];int a[305][305];int max(int a,int b){return a>b?a:b;}int main(){int n,i,j,k;scanf("%d",&n);for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&a[i][j]);int tol=2*(n-1);for(i=0;i<=tol;i++)for(j=0;j<=n;j++)for(k=0;k<=n;k++)dp[i][j][k]=-INF;//初值要设成负的，因为有负值dp[0][0][0]=0;for(i=0;i<tol;i++)for(j=0;j<=i&&j<=n-1;j++)for(k=0;k<=i&&k<=n-1;k++){int x1=i-j+1,y1=j+1,x2=i-k+1,y2=k+1;//当前所在位置if(y1+1<=n&&x2+1<=n)//1往右2往下{int nx1=x1,ny1=y1+1,nx2=x2+1,ny2=y2;int tmp;if(nx1==nx2&&ny1==ny2)tmp=a[nx1][ny1];elsetmp=a[nx1][ny1]+a[nx2][ny2];dp[i+1][j+1][k]=max(dp[i+1][j+1][k],dp[i][j][k]+tmp);}if(x1+1<=n&&y2+1<=n)//1往下2往右{int nx1=x1+1,ny1=y1,nx2=x2,ny2=y2+1;int tmp;if(nx1==nx2&&ny1==ny2)tmp=a[nx1][ny1];elsetmp=a[nx1][ny1]+a[nx2][ny2];dp[i+1][j][k+1]=max(dp[i+1][j][k+1],dp[i][j][k]+tmp);}if(x1+1<=n&&x2+1<=n)//1往下2往下{int nx1=x1+1,ny1=y1,nx2=x2+1,ny2=y2;int tmp;if(nx1==nx2&&ny1==ny2)tmp=a[nx1][ny1];elsetmp=a[nx1][ny1]+a[nx2][ny2];dp[i+1][j][k]=max(dp[i+1][j][k],dp[i][j][k]+tmp);}if(y1+1<=n&&y2+1<=n)//1往右2往右{int nx1=x1,ny1=y1+1,nx2=x2,ny2=y2+1;int tmp;if(nx1==nx2&&ny1==ny2)tmp=a[nx1][ny1];elsetmp=a[nx1][ny1]+a[nx2][ny2];dp[i+1][j+1][k+1]=max(dp[i+1][j+1][k+1],dp[i][j][k]+tmp);}}printf("%d\n",dp[tol][n-1][n-1]+a[1][1]);return 0;}