悬 线 法

悬线法：求满足某条件的极大子矩阵，比如“面积最大的长方形、正方形”“周长最长的矩形等等”。
一般用于求 n ✖ m 的01矩阵中面积最大全为1的子矩阵的面积，时间复杂度O(nm)
单调栈求最大子矩阵
1
2

悬线法板子
滚动数组优化

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 2e3 + 9;
int n, m, i, j, ans;
int l[maxn], r[maxn], h[maxn];
int lmax, rmax, a[maxn][maxn];
int main()
{
	for(scanf("%d %d", &n, &m), i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j)
			scanf("%d", &a[i][j]);
	for(int i = 1; i <= m; ++i)	l[i] = 1, r[i] = m;
	for(int i = 1; i <= n; ++i)
	{
		for(lmax = j = 1; j <= m; ++j)
		{
			if(a[i][j])	h[j]++, l[j] = max(l[j], lmax);
			else h[j] = 0, l[j] = 1, r[j] = m, lmax = j + 1;
		}
		for(rmax = j = m; j; --j)
		{
			if(a[i][j])
			{
				r[j] = min(r[j], rmax);
				ans = max(ans, (r[j] - l[j] + 1) * h[j]);
			}
			else rmax = j - 1;
		}
	}
	printf("%d\n", ans);
	return 0;
}
/*
4 4
0 0 0 0
1 1 1 1
1 0 1 1
1 1 1 1
*/

例题
P1169-ZJOI2007-棋盘制作

#include <iostream>
#include <algorithm>
using namespace std;
 
const int Max = 2005;
 
int ves[Max][Max], up[Max][Max], Left[Max][Max], Right[Max][Max];
int temp1 = 1, temp2 = 1;

int main()
{
	ios::sync_with_stdio(false);
	int N, M;
	cin >> N >> M;
	for(int i = 1; i <= N; i++)
		for (int j = 1; j <= M; j++) {
			cin >> ves[i][j];
			Left[i][j] = Right[i][j] = j;	//初始化Right和Left，使他们值为点所在纵坐标
			up[i][j] = 1;	//初始化up使其值为1
		}
 
	for (int i = 1; i <= N; i++)
		for (int j = 2; j <= M; j++)
			if (ves[i][j] == 1 - ves[i][j - 1])	//判断相邻两个数是否不同
				Left[i][j] = Left[i][j - 1];	// 不同则i可以向左拓展一个 
 
	for (int i = 1; i <= N; i++)
		for (int j = M - 1; j > 0; j--)
			if (ves[i][j] == 1 - ves[i][j + 1])
				Right[i][j] = Right[i][j + 1];// 右边界可以向右拓展一个 

	for(int i = 1;i <= N; i++)
		for (int j = 1; j <= M; j++) {
			if (i > 1 && ves[i][j] == 1 - ves[i - 1][j]) {	//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]);
				up[i][j] = up[i - 1][j] + 1;
			}// 对齐边界 
			int A_instance = Right[i][j] - Left[i][j] + 1;	//计算长度
			int B_instance = min(A_instance, up[i][j]);	//算出长宽中较小的边，以计算正方形
			temp1 = max(temp1, B_instance * B_instance);	//正方形面积
			temp2 = max(temp2, A_instance * up[i][j]);		//长方形面积
		}
	cout << temp1 << endl << temp2 << endl;
	return 0;
}

hdu-6957-Maximal submatrix
思路：转化成01矩阵悬线法求解
这个题要注意ans初始化成m，还有就是求矩阵面积时高度要+1（因为第一层没有标记成1，但是也不能标记，一旦第一层标记就会导致混乱了）

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 2e3 + 9;
int n, m, i, j, ans;
int l[maxn], r[maxn], h[maxn];
int lmax, rmax, a[maxn][maxn], b[maxn][maxn];
void work()
{
	
	for(scanf("%d %d", &n, &m), i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j)
			scanf("%d", &b[i][j]), a[i][j] = 0;
	for(i = 2; i <= n; ++i)//  从第二层开始建立01矩阵
		for(j = 1; j <= m; ++j)
			if(b[i][j] >= b[i-1][j]) a[i][j] = 1;
	ans = m;//  一定要记得初始化成 m 
	for(int i = 1; i <= m; ++i)	l[i] = 1, r[i] = m, h[i] = 0;
	for(int i = 1; i <= n; ++i)
	{
		for(lmax = j = 1; j <= m; ++j)
			if(a[i][j])	h[j]++, l[j] = max(l[j], lmax);
			else h[j] = 0, l[j] = 1, r[j] = m, lmax = j + 1;
		for(rmax = j = m; j; --j)
			if(a[i][j])
			{
				r[j] = min(r[j], rmax);
				ans = max(ans, (r[j] - l[j] + 1) * (h[j] + 1));
			}
			else rmax = j - 1;
	}
	printf("%d\n", ans);
}
int main()
{
	int T;cin >> T;while(T--)
		work();
	return 0;
}