[dp] Codeforces 429B B. Working out

题目链接：

http://codeforces.com/problemset/problem/429/B

• PROBLEMS
• SUBMIT
• STATUS
• STANDINGS
• CUSTOM TEST

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.

Sample test(s)

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].

题目意思：

给n*m的矩阵，每个格子有个数，A从(1,1)出发只能向下或右走，终点为(n,m)，B从(n,1)出发只能向上或右走，终点为(1,m)。两个人的速度不一样，走到的格子可以获的该格子的数，两人相遇的格子上的数两个人都不能拿。求A和B能拿到的数的总和的最大值。

n,m<=1000

解题思路：

dp.

先预处理出每个格子到四个角落格子的路径最大数值，然后枚举两个人相遇的交点格子，枚举A、B的进来和出去方式，求最大值即可。

注意边界情况。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 1100
ll dp[5][Maxn][Maxn];
int n,m;

ll save[Maxn][Maxn];


int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   while(~scanf("%d%d",&n,&m))
   {
       memset(save,0,sizeof(save));

       for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%d",&save[i][j]);

       memset(dp,0,sizeof(dp));
       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])+save[i][j];
               //printf("i:%d j:%d %I64d\n",i,j,dp[1][i][j]);
           for(int j=m;j>=1;j--)
                dp[3][i][j]=max(dp[3][i-1][j],dp[3][i][j+1])+save[i][j];
       }

       for(int i=n;i>=1;i--)
       {
           for(int j=1;j<=m;j++)
                dp[2][i][j]=max(dp[2][i+1][j],dp[2][i][j-1])+save[i][j];
           for(int j=m;j>=1;j--)
                dp[4][i][j]=max(dp[4][i+1][j],dp[4][i][j+1])+save[i][j];
       }
       ll ans=0;
       for(int i=0;i<=n+1;i++)  //把四个边界都置为无效情况
            for(int k=1;k<=4;k++)
                dp[k][i][0]=dp[k][i][m+1]=-INF;
       for(int j=0;j<=m+1;j++)
            for(int k=1;k<=4;k++)
                dp[k][0][j]=dp[k][n+1][j]=-INF;

       for(int i=1;i<=n;i++)
       {
           for(int j=1;j<=m;j++)
           {     //只用考虑两种情况 两个人只能交叉一个格子，多了的话不合算
               ll temp=dp[1][i][j-1]+dp[4][i][j+1]+dp[2][i+1][j]+dp[3][i-1][j];
               ans=max(ans,temp);
               temp=dp[1][i-1][j]+dp[4][i+1][j]+dp[2][i][j-1]+dp[3][i][j+1];
               ans=max(ans,temp);
               //printf("i:%d j:%d %I64d\n",i,j,ans);
           }
       }
       printf("%I64d\n",ans);
   }
   return 0;
}