CodeForces 173C Spiral Maximum

题意：给你一个n*m的矩阵，然后问你螺旋线能够覆盖的最大和是多少
思路：
暴力记忆化搜索，滚动数组优化
dp[i][j][len]表示以i,j为起点，正方形边长为len的覆盖的值是多少
dp[i][j][len]显然等于len所在的正方形覆盖的和 - mp[i+1][j] - dp[i+1][j+1][len-2]

#include<bits/stdc++.h>
using namespace std;
#define INF 1e9
int n,m;
int mapp[510][510];
int sum[510][510];
int vis[510][510];
int dp[510][510][2];
int cal(int x1,int y1,int x2,int y2)
{
return sum[x2][y2]-sum[x1-1][y2]-sum[x2][y1-1]+sum[x1-1][y1-1];
}
int solve(int x,int y,int len)
{
if (len==0)
{
dp[x][y][0]=mapp[x][y];
return dp[x][y][0];
}
else
{
dp[x][y][0]=cal(x,y,x+len,y+len);
dp[x][y][0]-=mapp[x+1][y];
dp[x][y][0]-=dp[x+1][y+1][1];
return dp[x][y][0];
}
}
int main()
{
scanf("%d%d",&n,&m);
for (int i = 1;i<=n;i++)
for (int j = 1;j<=m;j++)
scanf("%d",&mapp[i][j]);
for (int i = 1;i<=n;i++)
for (int j = 1;j<=m;j++)
sum[i][j]=mapp[i][j]+sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];
int ans = -INF;
for (int k = 0;k<=min(n,m);k+=2)
{
memset(vis,0,sizeof(vis));
for (int i = n;i>=1;i--)
for (int j = m;j>=1;j--)
dp[i][j][1]=dp[i][j][0];
for (int i = n;i>=1;i--)
{
if (i+k>n)
continue;
for (int j = m;j>=1;j--)
{
if (j+k>m)
continue;
int p = solve(i,j,k);
if (k>0)
ans = max(ans,p);
}
}
}
printf("%d\n",ans);
}

Description

Let's consider a k × k square, divided into unit squares. Please note that k ≥ 3 and is odd. We'll paint squares starting
from the upper left square in the following order: first we move to the right, then down, then to the left, then up, then to the right again and so on. We finish moving in some direction in one of two cases: either we've reached the square's border or the
square following after the next square is already painted. We finish painting at the moment when we cannot move in any direction and paint a square. The figure that consists of the painted squares is a spiral.


The figure shows examples of spirals for k = 3, 5, 7, 9.
You have an n × m table, each of its cells contains a number. Let's consider all possible spirals, formed by the table cells. It means that we consider all spirals of any size that
don't go beyond the borders of the table. Let's find the sum of the numbers of the cells that form the spiral. You have to find the maximum of those values among all spirals.

Input

The first line contains two integers n and m (3 ≤ n, m ≤ 500)
— the sizes of the table.

Each of the next n lines contains m space-separated integers: the j-th
number in the i-th line aij ( - 1000 ≤ aij ≤ 1000)
is the number recorded in the j-th cell of the i-th row of the table.

Output

Print a single number — the maximum sum of numbers among all spirals.

Sample Input

Input

6 5
0 0 0 0 0
1 1 1 1 1
0 0 0 0 1
1 1 1 0 1
1 0 0 0 1
1 1 1 1 1

Output

17

Input

3 3
1 1 1
1 0 0
1 1 1

Output

6

Input

6 6
-3 2 0 1 5 -1
4 -1 2 -3 0 1
-5 1 2 4 1 -2
0 -2 1 3 -1 2
3 1 4 -3 -2 0
-1 2 -1 3 1 2

Output

13

Hint

In the first sample the spiral with maximum sum will cover all 1's of the table.

In the second sample the spiral may cover only six 1's.