悬线法练习

在此感谢那个网名叫顾z的人。

悬线法：

目标：求给定矩阵中满足条件的最大子矩阵

方法：left[i][j]表示(i,j)所能到达的最左位置 , right[i][j]类似 , up[i][j]表示往上扩展的最大长度。

递推公式： left[i][j]=max(left[i][j],left[i−1][j]) right[i][j]=min(right[i][j],right[i-1][j])

正确性显然， 适用性可能不是那么通用。

讲解更好的博客：https://www.luogu.com.cn/blog/RPdreamer/p1169

//悬线法求解最大规模子矩阵板子

int main()
{
	
	cin >> n >> m;
	for(int i=1; i<=n; ++i)
		for(int j=1; j<=m; ++j)
		{
			cin >> a[i][j];
			left[i][j] = right[i][j] = j;
			up[i][j] = 1;
		}
	
	for(int i=1; i<=n; ++i)
		for(int j=2; j<=n; ++j)
			if(……)
				left[i][j] = left[i][j-1];
	
	for(int i=1; i<=n; ++i)
		for(int j=n-1; j>=1; --j)
			if(……)
				right[i][j] = right[i][j+1];
	
	//处理up=1时的right与left数组
	
	for(int i=1; i<=n; ++i)
		for(int j=1; j<=m; ++j)
		{
			
			if(i>1 && ……)
			{
				
				left[i][j] = max(left[i][j], left[i-1][j]);
				right[i][j] = min(right[i][j], right[i-1][j]);
				up[i][j] = up[i-1][j] + 1;
			}
			
			/* calc ans */
			
			
			
		}
		
	cout << ans;
		
	
	return 0;
	
 } 

---

例题：https://www.luogu.com.cn/problem/P1387

#include<bits/stdc++.h>
using namespace std;
const int maxn = 105;

int n, m, a[maxn][maxn];
int up[maxn][maxn], l[maxn][maxn], r[maxn][maxn];

int main()
{
	
	scanf("%d%d", &n, &m);
	for(int i=1; i<=n; ++i)
		for(int j=1; j<=m; ++j)
		{
			scanf("%d", &a[i][j]);
			l[i][j] = r[i][j] = j;
			up[i][j] = 1;
		}
		
	for(int i=1; i<=n; ++i)
		for(int j=2; j<=m; ++j)
			if(a[i][j-1] && a[i][j])
				l[i][j] = l[i][j-1];
	
	for(int i=1; i<=n; ++i)
		for(int j=m-1; j>=1; --j)
			if(a[i][j+1] && a[i][j])
				r[i][j] = r[i][j+1];
	
	int ans = 0;
	
	for(int i=1; i<=n; ++i)
		for(int j=1; j<=m; ++j)
		if(a[i][j] == 1)
		{
			
			if(i>1 && a[i-1][j] && a[i][j])
			{
				l[i][j] = max(l[i][j], l[i-1][j]);
				r[i][j] = min(r[i][j], r[i-1][j]);
				up[i][j] = up[i-1][j] + 1;
			}
			
			int a = r[i][j] - l[i][j] + 1;
			int b = up[i][j];
			ans = max(ans, min(a,b));
			
		}
	
	cout << ans;
	
	return 0;
	
}

例题：https://www.luogu.com.cn/problem/P1169

#include<bits/stdc++.h>
using namespace std;


const int maxn = 2005;

int n, m, a[maxn][maxn];

int l[maxn][maxn], r[maxn][maxn], up[maxn][maxn];

int main()
{
	
	cin >> n >> m;
	for(int i=1; i<=n; ++i)
		for(int j=1; j<=m; ++j)
		{
			scanf("%d", &a[i][j]);
			l[i][j] = r[i][j] = j;
			up[i][j] = 1;
		}
		
	for(int i=1; i<=n; ++i)
		for(int j=2; j<=m; ++j)
			if(a[i][j] != a[i][j-1])
				l[i][j] = l[i][j-1];
	
	for(int i=1; i<=n; ++i)
		for(int j=m-1; j>=1; --j)
			if(a[i][j] != a[i][j+1])
				r[i][j] = r[i][j+1];
	
	int ans1 = 0, ans2 = 0;
	
	for(int i=1; i<=n; ++i)
		for(int j=1; j<=m; ++j)
		{
			
			if(i>1 && a[i][j] != a[i-1][j])
			{
				l[i][j] = max(l[i][j], l[i-1][j]);
				r[i][j] = min(r[i][j], r[i-1][j]);
				up[i][j] = up[i-1][j] + 1;
			}
			
			int a = r[i][j] - l[i][j] + 1;
			int b = up[i][j];
			
			ans1 = max(ans1, min(a,b)*min(a,b) );
			ans2 = max(ans2, a*b);
			
		}
	
	cout << ans1 << '
' << ans2;
	
	return 0;
	
}