codeforces 729

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

 

题目描述：

 

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.

Examples

input

3 3 100 100 100 100 1 100 100 100 100

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：

 

给你一个n*m的方格，每个点都有其对应的价值，甲从（1，1）出发只能向右或向下走到(n,m)，乙从(n,1)出发只能向上或向右走到(1,m)，两者在中间有且仅有一个相遇的点，甲乙会收取走过的那个点的价值，但是相遇的那个点两个人都不拾取，求出取得最大的价值和

 

题目分析：

dp，分别对每个点的上下左右dp一下，详见代码中的注释

 

AC代码：

 

#include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; #define ll long long const int maxn=1005; ll dp[5][maxn][maxn],v[maxn][maxn]; int n,m; ll ans; int main() { while(cin>>n>>m) { memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) scanf("%lld",&v[i][j]); } //dp[1][i][j]代表从左上角到v[i][j]的最大值 for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) dp[1][i][j]=max(dp[1][i-1][j],dp[1][i][j-1])+v[i][j]; } //dp[2][i][j]代表从左下角到v[i][j]的最大值 for(int i=n;i>=1;i--) { for(int j=1;j<=m;j++) dp[2][i][j]=max(dp[2][i][j-1],dp[2][i+1][j])+v[i][j]; } //dp[3][i][j]代表从右上角到v[i][j]的最大值 for(int i=1;i<=n;i++) { for(int j=m;j>=1;j--) dp[3][i][j]=max(dp[3][i][j+1],dp[3][i-1][j])+v[i][j]; } //dp[4][i][j]代表从右下角到v[i][j]的最大值 for(int i=n;i>=1;i--) { for(int j=m;j>=1;j--) dp[4][i][j]=max(dp[4][i][j+1],dp[4][i+1][j])+v[i][j]; } //枚举相遇的点的位置，由于两条路径有且仅有一个交点，所以可能的路径也就只有两种，一种是在[i][j]点处从左上角到右下角的那条路径是从上到下的，另一条是从左到右的；第二种是从左上角到右上角是从左到右的，另一条的从下到上的 ans=0; for(int i=2;i<n;i++) { for(int j=2;j<m;j++) { ans=max(ans,dp[1][i-1][j]+dp[4][i+1][j]+dp[2][i][j-1]+dp[3][i][j+1]); ans=max(ans,dp[1][i][j-1]+dp[4][i][j+1]+dp[2][i+1][j]+dp[3][i-1][j]); } } cout<<ans<<endl; } return 0; }