1.【題目連結】

2.思路

記m = ∑^w_i

對於n <= 300的情況，暴力O(n^2)列舉矩形的上下邊界，O(m)計算矩形內的答案。

對於n > 300的情況，暴力O(n)列舉矩形的上邊界，然後列舉暴力O(m)列舉高度（最大全0子矩陣單調棧做法的高度），最後對於每個高度，O(m)計算矩形內的答案。

計算答案的過程是：

列舉每一個矩形，計算出所有矩形左邊界和右邊界能向內延伸的最大值和次大值，記錄次大值的原因是，如果左邊界右邊界最大值的矩形是同一個矩形，那麼就不合法了，就只能用次大值去更新答案。

3.Code

#include<bits/stdc++.h>
 
     //萬能標頭檔案
using namespace std;        //養成好習慣
 
const int maxn = 305, maxm = 100005;
 
int s, n, m, pos[maxm];
bitset<maxm> g[maxn], mp;
 
inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' &&
 
 ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}
 
inline int cread() {
	char ch = getchar(); for(; ch != '0' && ch != '1'; ch = getchar());
	return ch - '0';
}
 
namespace subtask1 {
	void solve() {
		int ans = 0;
		for(int i = 1; i <= s; i++) {  //for迴圈
			pos[i] = pos[i -
 
 1] + iread();
			for(int j = 1; j <= n; j++) for(int k = pos[i - 1] + 1; k <= pos[i]; k++)
				g[j][k] = cread();
		}
		m = pos[s];
		for(int i = 1; i <= n; i++) {  //for迴圈
			mp = g[i];
			for(int j = i; j <= n; j++) {   //for迴圈
				mp |= g[j];
				for(int k = 1, sum = 0; k <= m; k++) {
					sum = mp[k] ? 0 : sum + 1;
					ans = max(ans, sum * (j - i + 1));
				}
				int l1 = 0, l2 = 0, lno = 0;
				int r1 = 0, r2 = 0, rno = 0;
				int sum = 0, full = 0;
				for(int k = 1; k <= s; k++) {   //for迴圈
					sum = 0;
					for(int l = pos[k - 1] + 1; l <= pos[k]; l++, sum++)
						if(mp[l]) break;
					if(sum == pos[k] - pos[k - 1]) {
						full += sum;
						continue;
					}
					if(sum > l1) l2 = l1, l1 = sum, lno = k;
					else if(sum > l2) l2 = sum;
					sum = 0;
					for(int r = pos[k]; r >= pos[k - 1] + 1; r--, sum++)
						if(mp[r]) break;
					if(sum > r1) r2 = r1, r1 = sum, rno = k;
					else if(sum > r2) r2 = sum;
				}
				ans = max(ans, (((lno == rno) ? max(l1 + r2, l2 + r1) : l1 + r1) + full) * (j - i + 1));
			}
		}
		printf("%d\n", ans);   //輸出
	}
}
 
namespace subtask2 {
	int f[maxm];
 
	void solve() {
		int ans = 0;
		for(int i = 1; i <= s; i++) {
			pos[i] = pos[i - 1] + iread();
			for(int k = 1; k <= n; k++) for(int j = pos[i - 1] + 1; j <= pos[i]; j++)
				g[j][k] = cread();
		}
		m = pos[s];
		for(int i = n; i; i--) {
			for(int j = 1; j <= m; j++) f[j] = g[j][i] ? 0 : f[j] + 1;
			for(int j = 1; j <= m; j++) {
				for(int k = 1, sum = 0; k <= m; k++) {
					sum = f[k] < f[j] ? 0 : sum + 1;
					ans = max(ans, f[j] * sum);
				}
				int l1 = 0, l2 = 0, lno = 0;
				int r1 = 0, r2 = 0, rno = 0;
				int sum = 0, full = 0;
				for(int k = 1; k <= s; k++) {
					sum = 0;
					for(int l = pos[k - 1] + 1; l <= pos[k]; l++, sum++)
						if(f[l] < f[j]) break;
					if(sum == pos[k] - pos[k - 1]) {
						full += sum;
						continue;
					}
					if(sum > l1) l2 = l1, l1 = sum, lno = k;
					else if(sum > l2) l2 = sum;
					sum = 0;
					for(int r = pos[k]; r >= pos[k - 1] + 1; r--, sum++)
						if(f[r] < f[j]) break;
					if(sum > r1) r2 = r1, r1 = sum, rno = k;
					else if(sum > r2) r2 = sum;
				}
				ans = max(ans, (((lno == rno ? max(l1 + r2, l2 + r1) : l1 + r1) + full) * f[j]));
			}
		}
		printf("%d\n", ans);     //輸出
	}
}
 
int main() {
	for(int T = iread(); T; T--) {      //for迴圈
		s = iread(), n = iread();
		if(n <= 300) subtask1::solve();
		else subtask2::solve();
	}
	return 0;    //不寫return 0,成績return 0