-
Notifications
You must be signed in to change notification settings - Fork 21
/
Copy pathfortmoo.cpp
85 lines (76 loc) · 2.09 KB
/
fortmoo.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
#include <iostream>
#include <string>
#include <utility>
#include <sstream>
#include <algorithm>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <bitset>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <unordered_set>
using namespace std;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define SORT(vec) sort(vec.begin(), vec.end())
#define INF 1000000010
#define LL_INF 4500000000000000000
#define LSOne(S) (S & (-S))
#define EPS 1e-9
#define A first
#define B second
#define mp make_pair
#define pb push_back
#define PI acos(-1.0)
#define ll long long
#define MOD (int)(2e+9+11)
#define SET(vec, val, size) for (int i = 0; i < size; i++) vec[i] = val;
#define SET2D(arr, val, dim1, dim2) F0R(i, dim1) F0R(j, dim2) arr[i][j] = val;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<ll> vl;
int grid[210][210];
int sum(int x1, int y1, int x2, int y2) {
return grid[x2+1][y2+1] - grid[x1][y2+1] - grid[x2+1][y1] + grid[x1][y1];
}
int main() {
freopen("fortmoo.in", "r", stdin);
freopen("fortmoo.out", "w", stdout);
int n, m; cin >> n >> m;
SET2D(grid, 0, 210, 210);
F0R(i, n) F0R(j, m) {
char c; cin >> c;
grid[i+1][j+1] = grid[i+1][j] + grid[i][j+1] - grid[i][j];
if (c == 'X') {
grid[i+1][j+1]++;
}
}
int best = 0;
F0R(i, n) {
FOR(j, i, n) {
int left = -1;
F0R(k, m) {
bool canUse = sum(i, k, j, k) == 0;
if (canUse) {
best = max(best, j-i+1);
}
if (left != -1 && canUse) {
best = max(best, (j-i+1)*(k-left+1));
}
if (sum(i, k, i, k) != 0 || sum(j, k, j, k) != 0) left = -1;
if (canUse && left == -1) left = k;
}
}
}
cout << best << endl;
return 0;
}