CodeForces 213C Relay Race(dp)

CodeForces 213C Relay Race

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.

题意：

接力赛跑，给出一张nxn的地图，一个人站在(1,1)，另一个人站在(n,n)，地图上每个点有一个数值(-1000~+1000)，先从(1,1)向(n,n)跑(只能向下或向右)，然后由(n,n)跑回(1,1)(只能向左或向上)，每跑到一个点会获得该点的数值，每个点的数值只能获得一次。

问能取得的数值的总和最大为多少。

分析：

相当于有两条由(1,1)到(n,n)的路径，求路径上所有点的值的总和最大为多少(每个点只能取一次)。d[time][x1][x2]，time为当前步数，x1为路径1的当前位置，x2为路径2的当前位置，由于每一步只跟上一步的状态有关，所以第一维可以简化大小为2。

考虑负数最大值情况初始化d。

#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>using namespace std;int max(int a,int b,int c){if(a>b){if(a>c)return a;else return c;}else{if(b>c)return b;else return c;}}int main(){int n,map[305][305],dp[2][305][305],x,ma;scanf("%d",&n);memset(dp,0,sizeof(dp));memset(map,0,sizeof(map));for(int i=0;i<2;i++){for(int j=0;j<305;j++){for(int k=0;k<305;k++){dp[i][j][k]=-100000000;}}}for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){scanf("%d",&map[i][j]);}}x=0;dp[0][1][1]=map[1][1];for(int i=1;i<2*n-1;i++){x^=1;for(int j=(i+1<n?1:i-n+2);j<=(i+1<n?i+1:n);j++){for(int k=j;k<=(i+1<n?i+1:n);k++){if(j==k){ma=max(dp[x^1][j-1][k-1],dp[x^1][j-1][k],dp[x^1][j][k]);dp[x][j][k]=ma+map[j][i-j+2];}else{ma=max(dp[x^1][j-1][k-1],dp[x^1][j-1][k],dp[x^1][j][k]);if(ma<dp[x^1][j][k-1])ma=dp[x^1][j][k-1];dp[x][j][k]=ma+map[j][i-j+2]+map[k][i-k+2];}//printf("j %d k %d dp %d\n",j,k,dp[x][j][k]);}}}if(n==1)printf("%d\n",map[1][1]);else printf("%d\n",dp[x][n][n]);return 0;}