Codeforce 题目429B Working out

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 3100 100 100100 1 100100 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 exercisesa[3][1] → a[2][1] → a[2][2] → a[2][3] → a[1][3].

题目大意：就是两个人，有一个矩阵，每个点都有一个val值，一个从左上角只能向右向下到右下角，一个人从左下角只能向上向右走到右上角，走到那个点获得一个那个点的val值，两个人路径肯定有相交的点，那个点能获得权值

思路：从四个角开始dp，求得走到每个出发角对应的角获得的dp值，然后枚举他们路径相交的点，然后再枚举一下走的方式就好了

ac代码

#include<stdio.h>#include<string.h>#include<iostream>#define max(a,b) (a>b?a:b)#define LL __int64#define INF 0x3f3f3f3fusing namespace std;LL dp[5][1010][1010];LL num[1010][1010];int main(){int n,m;while(scanf("%d%d",&n,&m)!=EOF){int i,j;memset(num,0,sizeof(num));for(i=1;i<=n;i++){for(j=1;j<=m;j++){scanf("%I64d",&num[i][j]);}}memset(dp,0,sizeof(dp));for(i=1;i<=n;i++){for(j=1;j<=m;j++){dp[1][i][j]=max(dp[1][i-1][j],dp[1][i][j-1])+num[i][j];}for(j=m;j>=1;j--){dp[2][i][j]=max(dp[2][i-1][j],dp[2][i][j+1])+num[i][j];}}for(i=n;i>=1;i--){for(j=1;j<=m;j++){dp[3][i][j]=max(dp[3][i+1][j],dp[3][i][j-1])+num[i][j];}for(j=m;j>=1;j--){dp[4][i][j]=max(dp[4][i+1][j],dp[4][i][j+1])+num[i][j];}}for(i=0;i<=n+1;i++){for(j=1;j<=4;j++)dp[j][i][0]=dp[j][i][m+1]=-INF;}for(i=0;i<=m+1;i++)for(j=1;j<=4;j++)dp[j][0][i]=dp[j][n+1][i]=-INF;LL ans=0;for(i=1;i<=n;i++){for(j=1;j<=m;j++){LL 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);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);}}printf("%I64d\n",ans);}}