Problem:
Largest Square

Given a matrix (H × W) which contains only 1 and 0, find the area of the largest square matrix which only contains 0s.

Input

H W
c1,1 c1,2 ... c1,W
c2,1 c2,2 ... c2,W
:
cH,1 cH,2 ... cH,W

In the first line, two integers H and W separated by a space character are given. In the following H lines, ci,j, elements of the H × W matrix, are given.

Output

Print the area (the number of 0s) of the largest square.

Constraints

1 ≤ H, W ≤ 1,400

Sample Input

4 5
0 0 1 0 0
1 0 0 0 0
0 0 0 1 0
0 0 0 1 0

Sample Output

4

Solution in C++14 (judge verdict: Accepted)

#include <iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<math.h>
using namespace std;
typedef long long ll;
#define int long long
typedef vector<int> VI;
typedef pair<int, int> pii;
typedef vector<pii> VP;
typedef vector<string> VS;
typedef priority_queue<int> PQ;
template<class T>bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }
#define fore(i,a) for(auto &i:a)
#define REP(i,n) for(int i=0;i<n;i++)
#define eREP(i,n) for(int i=0;i<=n;i++)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define eFOR(i,a,b) for(int i=(a);i<=(b);++i)
#define SORT(c) sort((c).begin(),(c).end())
#define rSORT(c) sort((c).rbegin(),(c).rend())
#define LB(x,a) lower_bound((x).begin(),(x).end(),(a))
#define UB(x,a) upper_bound((x).begin(),(x).end(),(a))
#define INF 1000000000
#define LLINF 9223372036854775807
#define mod 1000000007
//priority_queue<int,vector<int>, greater<int> > q2;



signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int H, W; cin >> H >> W;
	vector<VI> A(H, VI(W));
	vector<VI> dp(H, VI(W, 0));
	REP(i, H)REP(j, W)cin >> A[i][j];
	REP(i, H)if (A[i][0] == 0)dp[i][0] = 1;
	REP(i, W)if (A[0][i] == 0)dp[0][i] = 1;

	REP(i, H - 1) {
		REP(j, W - 1) {
			if (A[i + 1][j + 1] == 0) {
				//cout << i << " " << j << endl;
				dp[i + 1][j + 1] = min({ dp[i][j],dp[i + 1][j],dp[i][j + 1] }) + 1;
			}
		}
	}
	
	int ans = 0;
	REP(i, H)REP(j, W)chmax(ans, dp[i][j]);
	cout << ans * ans << endl;
	return 0;
}