Lưu ý: Các code mẫu dưới đây chỉ mang tính tham khảo và có thể không AC được bài tập này

Code mẫu của flashmt

const fi='';
      fo='';
      maxn=1010;
var n,re,m,last:longint;
    a,h,hh:array[0..1,0..maxn] of longint;
    st,d:array[0..maxn] of longint;

procedure rf;
var i,j:longint;
begin
     read(m,n);
     st[0]:=-maxlongint;
     for i:=0 to 1 do
     begin
          a[i,n+1]:=maxlongint;
          a[i,0]:=maxlongint;
     end;
     for j:=1 to n do a[0,j]:=maxlongint;
end;

procedure pr;
var i,j,z,t:longint;
begin
     for i:=1 to m do
     begin
          last:=0; z:=i and 1;
          for j:=1 to n do
          begin
               read(a[z,j]);
               if a[z,j]>=a[1-z,j] then
               begin
                    h[z,j]:=h[1-z,j]+1;
                    hh[z,j]:=hh[1-z,j]+1;
               end
               else
               begin
                    h[z,j]:=1;
                    hh[z,j]:=1;
               end;
               if a[1-z,j]>a[1-z,j+1] then hh[z,j]:=1;
               if h[z,j]>re then re:=h[z,j];
               if a[z,j]<a[z,j-1] then
               begin
                    last:=1; st[1]:=hh[z,j]; d[1]:=j;
                    continue;
               end;
               if hh[z,j]>st[last] then
               begin
                    inc(last); st[last]:=hh[z,j]; d[last]:=j;
                    continue;
               end
               else
               begin
                    while hh[z,j]<=st[last] do
                    begin
                         if st[last]<=h[z,j] then t:=1 else t:=0;
                         if re<st[last]*(j-d[last]+t) then
                            re:=st[last]*(j-d[last]+t);
                         dec(last);
                    end;
                    inc(last);
                    if hh[z,j]<>st[last] then st[last]:=hh[z,j];
               end;
          end;
          for j:=1 to last do
              if re<st[j]*(n-d[j]+1) then
                 re:=st[j]*(n-d[j]+1);
     end;
     writeln(re);
end;

begin
     assign(input,fi); reset(input);
     assign(output,fo); rewrite(output);
     rf;
     pr;
     close(input); close(output);
end.

Code mẫu của ladpro98

#include <cstring>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <climits>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <cassert>
#include <climits>
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define REP(i, a, b) for(int i = (a); i <=(b); i++)
#define FORD(i, a, b) for(int i = (a); i > (b); i--)
#define REPD(i, a, b) for(int i = (a); i >=(b); i--)
#define TR(it, a) for(typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define RESET(a, v) memset((a), (v), sizeof((a)))
#define SZ(a) (int((a).size()))
#define ALL(a) (a).begin(), (a).end()
#define PB push_back
#define MP make_pair
#define LL long long
#define LD long double
#define II pair<int, int>
#define X first
#define Y second
#define VI vector<int>

const int N = 1010;

using namespace std;

int a[N][N], L[N], R[N], h[N], lcp[N];
int n, m, ans;

int main() {
    ios :: sync_with_stdio(0); cin.tie(0);
    cin >> m >> n;
    REP(i, 1, m) REP(j, 1, n) cin >> a[i][j];
    REP(i, 1, m) {
        REP(j, 1, n)
        if (a[i][j] >= a[i - 1][j]) h[j]++;
            else h[j] = 1;
        REP(j, 2, n)
        if (a[i][j] >= a[i][j - 1]) lcp[j]++;
            else lcp[j] = 0;
        REP(j, 2, n)
            lcp[j] = min(lcp[j], min(h[j - 1], h[j]));
        REP(j, 2, n)
        for(L[j] = j - 1; L[j] > 1 && lcp[L[j]] >= lcp[j]; L[j] = L[L[j]]);
        REPD(j, n, 2)
        for(R[j] = j + 1; R[j] <= n && lcp[R[j]] >= lcp[j]; R[j] = R[R[j]]);
        REP(j, 1, n) ans = max(ans, h[j]);
        REP(j, 2, n) ans = max(ans, lcp[j] * (R[j] - L[j]));
    }
    cout << ans;
    return 0;
}

Code mẫu của RR

{$R+,Q+}
PROGRAM NKGOLF;
CONST
  FINP='';
  FOUT='';
  maxn=1005;
  oo=1000000;
VAR
  m,n,kq:longint;
  a:array[0..maxn,0..maxn] of longint;
  d2,d1:array[0..maxn,0..maxn] of longint;
  stack,pre:array[1..maxn*maxn] of longint;
  top:longint;
procedure readInput;
var
  f:text;
  i,j:longint;
begin
  assign(f,FINP); reset(f);
  read(f,m,n);
  for i:=1 to m do
    for j:=1 to n do
      read(f,a[i,j]);
  close(f);
end;
procedure solve1;
var
  i,j:longint;
begin
  for j:=1 to n do
    d1[1,j]:=1;
  kq:=0;
  for i:=2 to m do
    for j:=1 to n do
      begin
        if a[i,j]>=a[i-1,j] then d1[i,j]:=d1[i-1,j]+1
        else d1[i,j]:=1;
        if d1[i,j]>kq then kq:=d1[i,j];
      end;
end;
procedure writeOutput;
var
  f:text;
begin
  assign(f,FOUT); rewrite(f);
  writeln(f,kq);
  close(f);
end;
procedure pop(var u,v:longint);
begin
  u:=stack[top]; v:=pre[top]; dec(top);
end;
procedure push(u,v:longint);
begin
  inc(top); stack[top]:=u; pre[top]:=v;
end;
function min(a,b:longint):longint;
begin
  if a<b then min:=a else min:=b;
end;
procedure findMax(i:longint);
var
  j,v,u,luu:longint;
begin
  top:=0;
  for j:=1 to n do
    begin
      v:=j;
      if top>0 then luu:=stack[top];
      while (top>0) and (d2[i,stack[top]]>=d2[i,j]) do
        begin
          pop(u,v);
          if v>1 then
          begin
            if (luu-v+2)*min(d2[i,u],d1[i,v-1])>kq then
              kq:=(luu-v+2)*min(d2[i,u],d1[i,v-1]);
          end
          else
            if (luu-v+1)*d2[i,u]>kq then kq:=(luu-v+1)*d2[i,u];
        end;
      push(j,v);
    end;
  if top>0 then luu:=stack[top];
  while top>0 do
    begin
      pop(u,v);
      if v>1 then
      begin
        if kq<(luu-v+2)*min(d2[i,u],d1[i,v-1])
          then kq:=(luu-v+2)*min(d2[i,u],d1[i,v-1]);
      end
      else if (luu-v+1)*d2[i,u]>kq then kq:=(luu-v+1)*d2[i,u];
    end;
end;
procedure solve2;
var
  i,j:longint;
begin
  for i:=1 to m do a[i,0]:=-oo;
  for j:=1 to n do a[0,j]:=-oo;
  for i:=1 to m do
    for j:=1 to n do
      if a[i,j]<a[i,j-1] then d2[i,j]:=0
      else
        if a[i,j]<a[i-1,j] then d2[i,j]:=1
        else d2[i,j]:=d2[i-1,j]+1;
  for i:=1 to m do
    findMax(i);
end;
BEGIN
  readInput;
  solve1;
  solve2;
  writeOutput;
END.

Code mẫu của hieult

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iomanip>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;

#define Rep(i,n) for(typeof(n) i = 0; i < (n); ++i)
#define Repd(i,n) for(typeof(n) i = (n)-1; i >= 0; --i)
#define For(i,a,b) for(typeof(b) i = (a); i <= (b); ++i)
#define Ford(i,a,b) for(typeof(a) i = (a); i >= (b); --i)
#define Fit(i,v) for(typeof((v).begin()) i = (v).begin(); i != (v).end(); ++i)
#define Fitd(i,v) for(typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++i)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define sz(a) ((int)(a).size())
#define all(a) (a).begin(), (a).end()
#define ms(a,x) memset(a, x, sizeof(a))
#define nl puts("")
#define sp printf(" ")
#define ok puts("ok")
//#include <conio.h>

template<class F, class T> T convert(F a, int p = -1) { stringstream ss; if (p >= 0) ss << fixed << setprecision(p); ss << a; T r; ss >> r; return r; }
template<class T> void db(T a, int p = -1) { if (p >= 0) cout << fixed << setprecision(p); cout << a << " "; }
template<class T> T gcd(T a, T b) { T r; while (b != 0) { r = a % b; a = b; b = r; } return a; }
template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<class T> T sqr(T x) { return x * x; }
template<class T> T cube(T x) { return x * x * x; }
template<class T> struct Triple { T x, y, z; Triple() {} Triple(T _x, T _y, T _z) : x(_x), y(_y), z(_z) {} };
template<class T> Triple<T> euclid(T a, T b) { if (b == 0) return Triple<T>(1, 0, a); Triple<T> r = euclid(b, a % b); return Triple<T>(r.y, r.x - a / b * r.y, r.z); }
template<class T> int getbit(T s, int i) { return (s >> i) & 1; }
template<class T> T onbit(T s, int i) { return s | (T(1) << i); }
template<class T> T offbit(T s, int i) { return s & (~(T(1) << i)); }
template<class T> int cntbit(T s) { return s == 0 ? 0 : cntbit(s >> 1) + (s & 1); }
const int bfsz = 1 << 16; char bf[bfsz + 5]; int rsz = 0;int ptr = 0;
char gc() { if (rsz <= 0) { ptr = 0; rsz = fread(bf, 1, bfsz, stdin); if (rsz <= 0) return EOF; } --rsz; return bf[ptr++]; }
void ga(char &c) { c = EOF; while (!isalpha(c)) c = gc(); }
int gs(char s[]) { int l = 0; char c = gc(); while (isspace(c)) c = gc(); while (c != EOF && !isspace(c)) { s[l++] = c; c = gc(); } s[l] = '\0'; return l; }
template<class T> bool gi(T &v) {
    v = 0; char c = gc(); while (c != EOF && c != '-' && !isdigit(c)) c = gc(); if (c == EOF) return false; bool neg = c == '-'; if (neg) c = gc();
    while (isdigit(c)) { v = v * 10 + c - '0'; c = gc(); } if (neg) v = -v; return true;
}

const double PI = 2 * acos(0);
const string months[] = {"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"};
const int days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int dr[] = {0, +1, 0, -1};
const int dc[] = {+1, 0, -1, 0};
const int inf = (int)1e9 + 5;
const ll linf = (ll)1e16 + 5;
const double eps = 1e-9;
const ll mod = 100000007;
typedef pair<int, int> II;

#define maxn 1005

int l[maxn], r[maxn], h[maxn], a[maxn][maxn], x[maxn][maxn];
int m,n,kq = 0;

int doc(){
    int res = 1, run;
    For(j, 1, n){
        run = 1;
        For(i, 2, m){
            if(x[i][j] >= x[i - 1][j]) run++;
            else run = 1;
            res = max(res, run);
        }
    }
    return res;
}

int ngang(){
    int res = 1, run;
    For(i, 1, m){
        run = 1;
        For(j, 2, n){
            if(x[i][j] >= x[i][j - 1]) run++;
            else run = 1;
            res = max(res, run);
        }
    }
    return res;
}

int cal(int r, int c){
    return (x[r][c] <= x[r + 1][c] && x[r][c] <= x[r][c + 1] && x[r + 1][c] <= x[r + 1][c + 1] && x[r][c + 1] <= x[r + 1][c + 1]);
}

int main(){
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);

     gi(m); gi(n);
     For(i, 1, m) For(j, 1, n) gi(x[i][j]);
     kq = max(doc(), ngang());
     m--; n--;
     For(i, 1, m) For(j, 1, n) a[i][j] = cal(i, j);
     memset(h,0,sizeof(h));
     For(i, 1, m) {
          For(j, 1, n) {
               if(a[i][j])
                     h[j]++;
               else h[j] = 0;
          }
          h[0] = -1; h[n+1] = -1;
          For(j, 1, n){
               l[j] = j;
               while(h[l[j]-1]>=h[j]) l[j] = l[l[j]-1];
          }
          Ford(j, n, 1){
               r[j] = j;
               while(h[r[j]+1]>=h[j]) r[j] = r[r[j]+1];
          }
          For(j, 1, n){
               if(h[j] > 0)
               kq = max(kq,(h[j] + 1) *(r[j]-l[j]+2));
              // cout << kq << endl;
          }
     }
     printf("%d",kq);
     //getch();
}

Code mẫu của khuc_tuan

uses math;
type integer = longint;
var
   tt, res, i, j, m, n : integer;
   a : array[1..1000,1..1000] of longint;
   b : array[1..1000,1..1000] of boolean;
   h, left, right : array[1..1000] of integer;
begin
     read( m, n);
     for i:=1 to m do for j:=1 to n do read(a[i,j]);
     for i:=1 to m-1 do for j:=1 to n-1 do b[i,j] :=  (a[i,j]<=a[i+1,j]) and (a[i,j]<=a[i,j+1]) and (a[i+1,j]<=a[i+1,j+1]) and (a[i][j+1]<=a[i+1][j+1]);
     res := 1;
     for i:=1 to m do begin
         tt := 1;
         for j:=2 to n do if a[i,j]>=a[i,j-1] then begin inc(tt); res := max( res, tt) end else tt:=1;

     end;
     for j:=1 to n do begin
         tt := 1;
         for i:=2 to m do if(a[i,j]>=a[i-1,j]) then begin inc(tt); res := max( res, tt) end else tt:=1;
     end;

     {for i:=1 to m-1 do begin for j:=1 to n-1 do write(b[i,j]:7); writeln end;}

     for i:=1 to m-1 do begin
         for j:=1 to n-1 do if b[i,j] then inc(h[j]) else h[j]:=0;
         for j:=1 to n-1 do begin
             tt := j-1;
             while(tt>0) and (h[tt]>=h[j]) do tt := left[tt];
             left[j] := tt;
         end;
         for j:=n-1 downto 1 do begin
             tt := j+1;
             while(tt<n) and (h[tt]>=h[j]) do tt := right[tt];
             right[j] := tt;
         end;
         for j:=1 to n-1 do if h[j]>0 then begin
              res := max( res, (h[j]+1) * (right[j] - left[j]));
         end;
     end;
     writeln(res);
end.