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.
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).
The output contains a single number — the maximum total gain possible.
3 3
100 100 100
100 1 100
100 100 100
800
Iahub will choose exercises a[1][1] → a[1][2] → a[2][2] → a[3][2] → a[3][3]. Iahubina will choose exercisesa[3][1] → a[2][1] → a[2][2] → a[2][3] → a[1][3].
/*/ 题意： 两个人在一块草地上散步 分别从map[1][1]-->map[n][m];map[n][1]-->map[1][m]; 两个人一个只能向右或者下移动，一个只能向左下移动，并且两人只相遇一次 每走一块地会获得相应的值 问他们两个人走到目的地能获得值的总和的最大的值是多少。
思路：
这个题目很有意思，虽然算是个水题，因为两个人只能有一点同时走，并且这点的值两人都不取，那这块地就可以以这个相遇点为中心分成四块。
上图：
以一个角的状态方程为例子 dp[i][j]=dp[i][j]+max(dp[i-1][j],dp[i][j-1]);
AC代码： /*/
#include"map"
#include"cmath"
#include"string"
#include"cstdio"
#include"vector"
#include"cstring"
#include"iostream"
#include"algorithm"
using namespace std;
typedef long long LL;
#define memset(x,y) memset(x,y,sizeof(x))
const int MX=1005;
LL dp1[MX][MX];
LL dp2[MX][MX];
LL dp3[MX][MX];
LL dp4[MX][MX];
int main() {
int m,n;
memset(dp1,0);
memset(dp2,0);
memset(dp3,0);
memset(dp4,0);
while(~scanf("%d%d",&n,&m)) {
for(int i=1; i<=n; i++) {
for(int j=1; j<=m; j++) {
scanf("%I64d",&dp1[i][j]);
dp2[i][j]=dp3[i][j]=dp4[i][j]=dp1[i][j];
}
}
for(int i=1; i<=n; i++) {
for(int j=1; j<=m; j++) {
dp1[i][j]+=max(dp1[i-1][j],dp1[i][j-1]);
}
}
for(int i=1; i<=n; i++) {
for(int j=m; j>=1; j--) {
dp2[i][j]+=max(dp2[i-1][j],dp2[i][j+1]);
}
}
for(int i=n; i>=1; i--) {
for(int j=m; j>=1; j--) {
dp3[i][j]+=max(dp3[i+1][j],dp3[i][j+1]);
}
}
for(int i=n; i>=1; i--) {
for(int j=1; j<=m; j++) {
dp4[i][j]+=max(dp4[i+1][j],dp4[i][j-1]);
}
}
LL ans=0;
for(int i=2; i<n; i++) {
for(int j=2; j<m; j++) {
ans=max(ans,dp1[i-1][j]+dp3[i+1][j]+dp2[i][j+1]+dp4[i][j-1]);
ans=max(ans,dp1[i][j-1]+dp3[i][j+1]+dp2[i-1][j]+dp4[i+1][j]);
}
}
printf("%I64d
",ans);
}
return 0;
}