描述
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 thei-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 workouta[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.
输入
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).
输出
The output contains a single number — the maximum total gain possible.
样例输入
3 3
100 100 100
100 1 100
100 100 100
800
提示
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].
题意；有两条路径从（1,1）到（n，m）和从（n， 1）到（1， m）求两条路的最大权值，并且两条路只能相遇一次。
思路：要保证只有一个格子重合,那么只可能是以下两种情况:
1) A向右走,相遇后继续向右走,而B向上走,相遇后继续向上走
2) A向下走,相遇后继续向下走,而B向右走,相遇后继续向右走
代码：
#include<stdio.h>
#include<string.h>
#define N 2100
#define max(a, b)(a > b ? a : b)
int dp1[N][N];//计算从（1,1） 到（i，j）的最大权值。
int dp2[N][N];//计算从（i,j） 到（n，m）的最大权值。
int dp3[N][N];//计算从（n,1） 到（i，j）的最大权值。
int dp4[N][N];//计算从（i,j） 到（1，m）的最大权值。
int a[N][N];
int main(void)
{
int i, j, m, n, ans;
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));
for(i = 1; i <= n; ++i)
for(j = 1; j <= m; ++j)
scanf("%d", &a[i][j]);
for(i = 1; i <= n; ++i)
for(j = 1; j <= m; ++j)
dp1[i][j] = max(dp1[i-1][j], dp1[i][j-1]) + a[i][j];
for(i = n; i >= 1; --i)
for(j = m; j >= 1; --j)
dp2[i][j] = max(dp2[i][j+1], dp2[i+1][j]) + a[i][j];
for(i = n; i >= 1; --i)
for(j = 1; j <= m; ++j)
dp3[i][j] = max(dp3[i][j-1], dp3[i+1][j]) + a[i][j];
for(i = 1; i <= n; ++i)
for(j = m; j >= 1; --j)
dp4[i][j] = max(dp4[i-1][j], dp4[i][j+1]) + a[i][j];
ans = 0;
for(i = 2; i < n; ++i)//注意i的取值范围。
for(j = 2; j < m; ++j)//注意j的取值范围。
{
ans = max(ans, dp1[i][j-1] + dp2[i][j+1] + dp3[i+1][j] + dp4[i-1][j]);//情况1。
ans = max(ans, dp1[i-1][j] + dp2[i+1][j] + dp3[i][j-1] + dp4[i][j+1]);//情况2。
}
printf("%d ", ans);
}