思路
可以利用悬线法，处理对于每个点在高度为 \(h\) 时的左右边界，然后随着高度增加，这个边界表示的范围一定是单调不增的，但是高度又在增加，所以一直取 \(max\) 就对了
最后注意输出答案的三倍
\(C++\) \(AC\) \(Code:\)
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e3 + 5;
int l[MAXN][MAXN], r[MAXN][MAXN], h[MAXN][MAXN]; //分别存放位于点 (i, j) 时的高度为 h 的 左右边界
char mapp[MAXN][MAXN];
int n, m;
void init() {for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j)l[i][j] = r[i][j] = j, h[i][j] = 1; // 初始化边界为自身，高度为1
}
int main() {cin >> n >> m;for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j)cin >> mapp[i][j];init();for (int i = 1; i <= n; ++i) {for (int j = 2; j <= m; ++j)if (mapp[i][j] == 'F' && mapp[i][j - 1] == 'F')l[i][j] = l[i][j - 1]; // 尝试向左扩展边界for (int j = m - 1; j >= 1; --j)if (mapp[i][j] == 'F' && mapp[i][j + 1] == 'F')r[i][j] = r[i][j + 1]; // 尝试向右扩展边界}int ans = 0;for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j) {if (mapp[i][j] == 'R')continue;if(mapp[i - 1][j] == 'F') { // 更新高度增加时的左右边界h[i][j] = h[i - 1][j] + 1;l[i][j] = max(l[i][j], l[i - 1][j]);r[i][j] = min(r[i][j], r[i - 1][j]);}ans = max(ans, (r[i][j] - l[i][j] + 1) * h[i][j]); // 求面积最大值}cout << ans * 3;return 0;
}