提交連結：CF 22B

題面：

B. Bargaining Table time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Bob wants to put a new bargaining table in his office. To do so he measured the office room thoroughly and drew its plan: Bob's office room is a rectangular roomn

 × m meters. Each square meter of the room is either occupied by some furniture, or free. A bargaining table is rectangular, and should be placed so, that its sides are parallel to the office walls. Bob doesn't want to change or rearrange anything, that's why all the squares that will be occupied by the table should be initially free. Bob wants the new table to sit as many people as possible, thus its perimeter should be maximal. Help Bob find out the maximum possible perimeter of a bargaining table for his office.

Input

The first line contains 2 space-separated numbers n andm (1 ≤ n, m ≤ 25) — the office room dimensions. Then there follown lines with m characters 0 or 1 each. 0 stands for a free square meter of the office room. 1 stands for an occupied square meter. It's guaranteed that at least one square meter in the room is free.

Output

Output one number — the maximum possible perimeter of a bargaining table for Bob's office room.

Examples Input

3 3
000
010
000

Output

8

Input

5 4
1100
0000
0000
0000
0000

Output

16

題意：

    0代表空閒位置，1代表佔有位置，給定一個01組成的區域，問空閒區域最大的周長為多少。

解題：

    因為資料量特別小，可以採用區域和的形式，O（1）知道左上角點和右下角點之間1的個數（通過四塊區域和可以直觀地運算得出），列舉左上角點和右下角點即可。

程式碼：

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
using namespace std;
int mapp[30][30],sum[30][30];
int main()
{
	int n,m,ans=0,tmp;
	char s[30];
	memset(sum,0,sizeof(sum));
	scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
	{
		  scanf("%s",s);
		  for(int j=1;j<=m;j++)
		    mapp[i][j]=s[j-1]-'0';
    }
    for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+mapp[i][j];
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			for(int k=i+1;k<=n;k++)
			{
				for(int p=j+1;p<=m;p++)
				{
                   tmp=sum[k][p]-sum[i][p]-sum[k][j]+sum[i][j];
				   if(tmp==0)
				   {
					   tmp=2*(k-i+p-j);
					   if(tmp>ans)
						   ans=tmp;
				   }
				}
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}