题目的大意是给定一个数组，有两个人在走，其中一个从（1,1) ->（n,m）另一个是从（n, 1）- > (1,m)在这期间他们会相遇一次，相遇的时候不加数组值，求两个人走到终点的时候最大和是多少。题目如下：
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 nlines 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].
要想使得两个人相遇就必须一个从上到下一个从左到右。
对于他们相遇的点在到达之前有四个方向。
所以从四个方向考虑：
dp1[i][j] = max(dp1[i-1][j], dp1[i][j-1]) + a[i][j];
dp2[i][j] = max(dp2[i+1][j], dp2[i][j+1]) + a[i][j];
dp3[i][j] = max(dp3[i+1][j], dp3[i][j-1]) + a[i][j];
dp4[i][j] = max(dp4[i-1][j], dp4[i][j+1]) + a[i][j];
代码如下：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#define pi acos(-1)
#define For(i, a, b) for(int (i) = (a); (i) <= (b); (i) ++)
#define Bor(i, a, b) for(int (i) = (b); (i) >= (a); (i) --)
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-10;
const ll mod = 1e9 + 7;
inline int read(){
int ret=0,f=0;char ch=getchar();
while(ch>'9'||ch<'0') f^=ch=='-',ch=getchar();
while(ch<='9'&&ch>='0') ret=ret*10+ch-'0',ch=getchar();
return f?-ret:ret;
}
int a[maxn][maxn];
int dp1[maxn][maxn];
int dp2[maxn][maxn];
int dp3[maxn][maxn];
int dp4[maxn][maxn];
int main(){
int n, m;
while(~scanf("%d%d",&n,&m)){
memset(dp1, 0, sizeof(dp1));
memset(dp2, 0, sizeof(dp2));
memset(dp3, 0, sizeof(dp3));
memset(dp4, 0, sizeof(dp4));
memset(a, 0, sizeof(a));
For(i, 1, n)
For(j, 1, m)
scanf("%d",&a[i][j]);
/*For(i, 1, n)
For(j, 1, m)
printf("%d%c",a[i][j], j == m ? '\n' : ' ');*/
//左上
For(i, 1, n)
For(j, 1, m)
dp1[i][j] = max(dp1[i - 1][j], dp1[i][j - 1]) + a[i][j];
//右下
Bor(i, 1, n)
Bor(j, 1, m)
dp2[i][j] = max(dp2[i + 1][j], dp2[i][j + 1]) + a[i][j];
//左下
Bor(i, 1, n)
For(j, 1, m)
dp3[i][j] = max(dp3[i + 1][j], dp3[i][j - 1]) + a[i][j];
//右上
For(i, 1, n)
Bor(j, 1, m)
dp4[i][j] = max(dp4[i - 1][j], dp4[i][j + 1]) + a[i][j];
int ans = 0;
For(i, 2, n - 1){
For(j, 2, m - 1){
ans = max(ans, dp1[i - 1][j] + dp2[i + 1][j] + dp3[i][j - 1] + dp4[i][j + 1]);
ans = max(ans, dp1[i][j - 1] + dp2[i][j + 1] + dp3[i + 1][j] + dp4[i - 1][j]);
}
}
printf("%d\n",ans);
}
return 0;
}继续弥补自己的不足。