CF_22B_BargainingTable

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 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 numbers n and m (1 ≤ n, m ≤ 25)
— the office room dimensions. Then there follow n 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 test(s)

input

3 3
000
010
000

output

8

input

5 4
1100
0000
0000
0000
0000

output

16

数据量很小穷举就可以了

数据量大的话还需要多处理处理情况（预处理）

看题目要认真

#include <iostream>
#include <stdio.h>
using namespace std;

const int M=27;
char ho[M][M];

int le(int ist,int ien,int jst,int jen)
{
int f=1;
for(int i=ist;i<=ien;i++)
for(int j=jst;j<=jen;j++)
if(ho[i][j]=='1')
{
f=0;
break;
}
return f;
}

int main()
{
//freopen("1.in","r",stdin);
int n,m;
int fr;
while(scanf("%d%d",&n,&m)!=EOF)
{
fr=0;
for(int i=1;i<=n;i++)
{
scanf("%s",ho[i]+1);
for(int j=1;j<=m;j++)
if(ho[i][j]=='0')
fr++;
}
cout<<"fr "<<fr<<endl;
int maxlen=0;
for(int ist=1;ist<=n;ist++)
for(int ien=ist;ien<=n;ien++)
for(int jst=1;jst<=m;jst++)
for(int jen=jst;jen<=m;jen++)
{
if((ien-ist+1)*(jen-jst+1)>fr-1)
continue;
int len=2*(ien-ist+jen-jst+2);
if(len<=maxlen)
continue;
//int maxn=maxlen;
maxlen=max(maxlen,len*le(ist,ien,jst,jen));
//if(maxn!=maxlen)
//cout<<ist<<" "<<ien<<" "<<jst<<" "<<jen<<endl;
}
printf("%d\n",maxlen);
}
return 0;
}