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  1. #pragma GCC target("avx2")
  2. #pragma GCC optimize("O3")
  3. #pragma GCC optimize("unroll-loops")
  4. #include <bits/stdc++.h>
  5. using namespace std;
  6.  
  7. // #define int long long
  8. #define FOR(i, l, r) for(int i = (l); i <= (r); i++)
  9. #define FOD(i, r, l) for(int i = (r); i >= (l); i--)
  10. #define fi first
  11. #define se second
  12. const int maxn = 400+ 10;
  13. const int mod  = 1e9 + 7;
  14.  
  15. int n, m, ans = 0;
  16. int a[maxn][maxn];
  17. int tot;
  18. int num[maxn * maxn];
  19. short last[maxn * maxn][maxn];
  20. short h[maxn][maxn];
  21. int get(int x) 
  22. {
  23.     return lower_bound(num + 1, num + 1 + tot, x) - num;
  24. }
  25.  
  26. signed main()
  27. {
  28.     ios_base::sync_with_stdio(0);
  29.     cin.tie(0);cout.tie(0);
  30.     cin >> n >> m;
  31.     FOR(i, 1, n) FOR(j, 1, m) 
  32.     {
  33.         cin >> a[i][j];
  34.         num[++tot] = a[i][j];
  35.     }
  36.  
  37.     sort(num + 1, num + 1 + tot);
  38.     tot = unique(num + 1, num + 1 + tot) - num - 1;
  39.     FOR(i, 1, n) FOR(j, 1, m) a[i][j] = get(a[i][j]);
  40.     FOR(i, 1, n) 
  41.     {
  42.         FOR(j, 1, m) 
  43.         {
  44.             h[j][j] = max(h[j][j], last[a[i][j]][j]);
  45.             last[a[i][j]][j] = i;
  46.             ans = max(ans, (i - h[j][j]));
  47.         }
  48.  
  49.         FOD(l, m, 1)
  50.         {
  51.             FOR(r, l + 1, m)
  52.             {
  53.                 h[l][r] = max(h[l][r], max(h[l + 1][r], max(h[l][r - 1], max(last[a[i][r]][l], last[a[i][l]][r]))));
  54.                 ans = max(ans, (r - l + 1) * (i - h[l][r]));
  55.             }
  56.         }
  57.     }
  58.  
  59.     cout << ans << "\n";
  60. }
Success #stdin #stdout 0.01s 5316KB
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https://ideone.com/XFwUAx
language:
C++14 (gcc 8.3)
created:
2 days ago
visibility:
public

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