P题1351-1399

合集 - 数据结构和算法(2)

1.P题1351-13992024-06-30

2.B题3051-30992024-06-30

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P1387 最大正方形

题目描述

在一个 $n\times m$ 的只包含 $0$ 和 $1$ 的矩阵里找出一个不包含 $0$ 的最大正方形，输出边长。

输入格式

输入文件第一行为两个整数 $n,m(1\leq n,m\leq 100)$，接下来 $n$ 行，每行 $m$ 个数字，用空格隔开，$0$ 或 $1$。

输出格式

一个整数，最大正方形的边长。

样例 #1

样例输入 #1

4 4
0 1 1 1
1 1 1 0
0 1 1 0
1 1 0 1

样例输出 #1

2

直接暴力枚举

#include <iostream>
#include <algorithm>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const ull N = 1e3 + 100;
const ull M = 1e3 + 200;
ull a[N][M];
ull prefix[N][M];
ull diff[N][M];
 
int main()
{
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 1; j <= m; ++j)
			cin >> a[i][j];
	}
 
	//预处理前缀和
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 1; j <= m; ++j)
		{
			prefix[i][j] = prefix[i - 1][j] + prefix[i][j - 1] - prefix[i - 1][j - 1] + a[i][j];
		}
	}
 
	//暴力枚举？//滑动窗口问题
	int minwidows = min(n, m);
 
	while (minwidows > 1)
	{
		for (int i = 1; i+minwidows-1 <= n; ++i)
		{
			for (int j = 1; j+minwidows-1 <= m; ++j)
			{
				ll tmp = prefix[i + minwidows - 1][j + minwidows - 1] - prefix[i - 1][j + minwidows - 1] - prefix[i + minwidows - 1][j - 1] + prefix[i - 1][j - 1];
				if (tmp == minwidows * minwidows)
				{
					cout << minwidows << '\n';
					return 0;
 
				}
 
			}
		}
		minwidows--;
	}
	cout << minwidows << '\n';
 
 
 
	return 0;
}

或者使用dp来做

#include <iostream>
#include <vector>
#include <map>
#include <numeric>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
 
const int MAX = 1e3 + 100;
ll a[MAX][MAX];
ll dp[MAX][MAX];
ll prefix[MAX][MAX];
ll ans;
 
 
int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 1; j <= m; ++j)
		{
			cin >> a[i][j];
		}
	}
 
	//dp  状态转移方程为 dp[i][j]=min(dp[i-1][j],dp[i][j-1],dp[i-1][j-1])+1;
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 1; j <= m; ++j)
		{	if (a[i][j] == 1)
			{
					dp[i][j] = min(min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
					ans = max(dp[i][j], ans);
			}
 
		}
	}
	cout << ans << "\n";
 
	return 0;
}