CF 429B B.Working out 四个角递推

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
Copy
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].

题意：给出n*m的方格，a从左上到右下，b从左下到右下，a只能往右往下走，b只能往右往上走，减去重合的一个交叉点值，求a和b走过的路径权值最大值

题解：分析交叉点的位置 肯定不在四个边上，否则会存在多个交叉点，当a b位于交叉点，只有俩种情况：

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

然后枚举四个角的dp值 

最后max求最大值

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=1005;
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(cin>>n>>m){
        if(n==0||m==0)
        {
            break;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                cin>>a[i][j];
            }
        }
        memset(dp1,0,sizeof(dp1));
        memset(dp2,0,sizeof(dp3));
        memset(dp3,0,sizeof(dp3));
        memset(dp4,0,sizeof(dp4));
        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])+a[i][j];//左上方
            }
        }
        for(int i=n;i>=1;i--){
            for(int j=m;j>=1;j--){
                dp2[i][j] = max(dp2[i+1][j],dp2[i][j+1])+a[i][j];//右下方
            }
        }
        for(int i=n;i>=1;i--){
            for(int j=1;j<=m;j++){
                dp3[i][j] = max(dp3[i+1][j],dp3[i][j-1])+a[i][j];//左下方
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=m;j>=1;j--){
                dp4[i][j] = max(dp4[i-1][j],dp4[i][j+1])+a[i][j];//右上方
            }
        }
        int ans=-1;
        for(int i=2;i<n;i++){
            for(int j=2;j<m;j++){
                ans = max(ans,dp1[i-1][j]+dp2[i+1][j]+dp3[i][j-1]+dp4[i][j+1]);//上 下 左 右
                ans = max(ans,dp1[i][j-1]+dp2[i][j+1]+dp3[i+1][j]+dp4[i-1][j]);//左 右 上 下
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}