Codeforces Beta Round #22 (Div. 2 Only)-B. Bargaining Table

原题链接

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 withm 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 3000010000

output

8

input

5 411000000000000000000

output

16

用dp[i][j]表示左上角为(1, 1)右下角为(i,j)的矩形里有多少个1, dp[i][j] = dp[i-1][j] + dp[i][j-1] + dp[i-1][j-1];如果(i,j)本身是1则dp[i][j]++;

遍历每个点(i,j)如果str[i][j] == '0'则把(i,j)表示矩阵的右下角枚举矩阵的左下角(i, h1)和右上角(h2, j)首先判断这个矩阵内1的数目 = dp[i][j] - dp[i][h1-1] - dp[h2-1][j] + dp[h2-1][h1-1]， 如果矩阵内没有1则可算矩阵的周长

#include <bits/stdc++.h>#define maxn 200005using namespace std;typedef long long ll;char str[30][30];int dp[30][30];int main(){//freopen("in.txt", "r", stdin);int n, m;        scanf("%d%d", &n, &m);    for(int i = 1; i <= n; i++)     scanf("%s", str[i]+1);    for(int i = 1; i <= n; i++)     for(int j = 1; j <= m; j++){     if(str[i][j] == '1')      dp[i][j] = 1;     dp[i][j] += dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1]; }int ans = 0;for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++){     if(str[i][j] == '1')      continue; for(int h1 = i-1; h1 >= 0; h1--){  for(int h2 = j-1; h2 >= 0; h2--){ int d = dp[i][j] - dp[h1][j] - dp[i][h2] + dp[h1][h2];    if(d == 0){    ans = max(ans, 2 * (i - h1 + j - h2));}if(str[i][h2] == '1') break;}if(str[h1][j] == '1') break;   } } printf("%d\n", ans);  return 0;}