CodeForces 22B Bargaining Table 01矩阵求最大矩形面积

B - Bargaining Table
Crawling in process...Crawling failedTime
Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

SubmitStatusPracticeCodeForces
22B

Description

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.

Sample Input

Input

3 3
000
010
000

Output

8

Input

5 4
1100
0000
0000
0000
0000

Output

16

可参照hdu1505的思路 http://blog.csdn.net/qiqijianglu/article/details/8557634

///不想再说什么了，hdu和poj的数据真心弱。
///这题和hdu1505一样的思路。
#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<memory.h>
using namespace std;
const int N=30;
char str

;
//int mat

;
int h

;
int l

;
int r

;
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(h,0,sizeof(h));
        for(int i=0;i<n;i++)
        scanf("%s",str[i]);
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        {
            if(str[i][j]=='0')
            h[i+1][j+1]=h[i][j+1]+1;
            else
            h[i+1][j+1]=0;
        }

        for(int i=1;i<=n;i++)
        {
            l[i][1]=1;
            r[i][m]=m;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=2;j<=m;j++)
            {
                int t=j;
                while(t>1&&h[i][j]<=h[i][t-1]&&h[i][j]!=0)///h[i][j]!=0很重要
                t=t-1;
                l[i][j]=t;
            }
        }
        for(int i=1;i<=n;i++)
        for(int j=m-1;j>=1;j--)
        {
            int t=j;
            while(t<m&&h[i][j]<=h[i][t+1]&&h[i][j]!=0)
            t=t+1;
            r[i][j]=t;
        }

        int maxn=0;
        for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            int s=(r[i][j]-l[i][j]+1+h[i][j])*2;
           //  cout<<l[i][j]<<" "<<r[i][j]<<" "<<"&"<<" "<<i<<" "<<j<<" "<<s<<" *"<<endl;
            if(s>maxn)
            maxn=s;
        }
        cout<<maxn<<endl;
    }
}