Petrozavodsk Winter Camp, Warsaw U, 2014, A The Carpet

一个地图上有若干障碍，问允许出现一个障碍的最大子矩形为多大？

最大子矩形改编

#include<bits/stdc++.h>
using namespace std;
#define rep(i, j, k) for (int i = int(j); i <= int(k); ++ i)
#define dwn(i, j, k) for (int i = int(j); i >= int(k); -- i)
const int N = ;
int n, m, l[N][], r[N][], h[N][];
char s[N][N];

int main() {
    scanf("%d%d", &n, &m);
    rep(i, , n) scanf("%s", s[i] + );
    int ans = ;
    // 枚举障碍不在悬线上
    // l[j][0]表示最左扩展（不能包含障碍），l[j][1]表示最右扩展（能包含障碍）
    rep(j, , m) l[j][] = l[j][] = j, r[j][] = r[j][] = m +  - j;
    rep(i, , n) {
        int lb0 = , lb1 = ;
        rep(j, , m) {
            if (s[i][j] == '.') {
                h[j][] ++; ans = max(ans, h[j][]);
                l[j][] = max(min(l[j][], j - lb0), min(l[j][], j - lb1));
                l[j][] = min(l[j][], j - lb0);
            }
            else {
                lb1 = lb0;
                lb0 = j;
                l[j][] = l[j][] = j;
                h[j][] = ;
            }
        }
        int rb0 = m + , rb1 = m + ;
        dwn(j, m, ) {
            if (s[i][j] == '.') {
                r[j][] = max(min(r[j][], rb0 - j), min(r[j][], rb1 - j));
                r[j][] = min(r[j][], rb0 - j);
                ans = max(ans, (l[j][] + r[j][] - ) * h[j][]);
                ans = max(ans, (l[j][] + r[j][] - ) * h[j][]);
                ans = max(ans, (l[j][] + r[j][] - ) * h[j][]);
            }
            else {
                rb1 = rb0;
                rb0 = j;
                r[j][] = r[j][] = m +  - j;
            }
        }
    }
    // 枚举障碍在悬线上
    // 这时候的l[j][0]表示没有障碍的悬线最左扩展，l[j][1]表示有障碍的悬线最左扩展
    rep(j, , m) l[j][] = l[j][] = j, r[j][] = r[j][] = m +  - j, h[j][] = h[j][] = ;
    rep(i, , n) {
        int lb = ;
        rep(j, , m) {
            if (s[i][j] == '.') {
                h[j][] ++; h[j][] ++;
                l[j][] = min(l[j][], j - lb);
                l[j][] = min(l[j][], j - lb);
            }
            else {
                l[j][] = min(l[j][], j - lb);
                h[j][] = h[j][] + ; h[j][] = ;
                l[j][] = j;
                lb = j;
            }
        }
        int rb = m + ;
        dwn(j, m, ) {
            if (s[i][j] == '.') {
                r[j][] = min(r[j][], rb - j);
                r[j][] = min(r[j][], rb - j);
            }
            else {
                r[j][] = min(r[j][], rb - j);
                r[j][] = m +  - j;
                rb = j;
            }
        }
        rep(j, , m) {
            ans = max(ans, h[j][] * (l[j][] + r[j][] - ));
        }
    }
    cout << ans << '\n';
}
/*
4 5
#.#..
....#
..#..
....#
*/