CodeForces - 22B Bargaining Table（暴力+dp）

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 room n × 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 numbersn and m (1 ≤ n, m ≤ 25) — the office room dimensions. Then there follow n lines with mcharacters 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.

Example

Input

3 3
000
010
000

Output

8

Input

5 4
1100
0000
0000
0000
0000

Output

16

 题意：0代表空闲位置，1代表家具。尽量时桌子的周长最大（桌子是矩形）。

   个人理解，是尽量多放桌子，让他们围成矩形，使值最大。

求出每个点为底的最大的高度（0),然后枚举右下角，再枚举矩形的高度，然后算取长度，进而算取周长即可。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,m;
char mapp[50][50];
int h[50][50];
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(h,0,sizeof(h));
        for(int i=0; i<n; i++)
            scanf("%s",mapp[i]);
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
            {
                if(mapp[i][j]=='1')continue;
                if(i==0||mapp[i-1][j]=='1')h[i][j]=1;
                else h[i][j]=h[i-1][j]+1;
            }
        int ans=-1;
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                for(int k=1; k<=h[i][j]; k++)
                {
                    int l=j;
                    while(l>=0&&h[i][l]>=k)
                        l--;
                    ans=max(ans,(j-l)+k);
                }
        printf("%d\n",ans*2);
    }
    return 0;
}