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

Code mẫu của RR

//Written by RR
{$R+,Q+}
{$Mode objfpc}
{$inline on}

uses math;
const
  FINP          =       '';
  FOUT          =       '';
  MAXN          =       1111;
  MAXL          =       111;
  oo            =       1000111000111000;
var
  f1,f2         :       text;
  l,w,n         :       longint;
  len,sum       :       array[0..MAXN] of longint;
  cost          :       array[1..MAXN,1..MAXN] of int64;
  f             :       array[0..MAXL,0..MAXN] of int64;
  test          :       longint;
  s,ss          :       ansistring;

procedure openF;
    begin
      assign(f1,FINP); reset(f1);
      assign(f2,FOUT); rewrite(f2);
    end;
procedure closeF;
    begin
      close(f1);
      close(f2);
    end;
procedure inp;
    var
      i:longint;
    begin
      readln(f1,l,w);
      readln(f1,s); n:=0;
      while (s<>'') do
        begin
          while s[1]=' ' do delete(s,1,1);
          while (s[length(s)]=' ') do delete(s,length(s),1);
          while pos('  ',s)>0 do delete(s,pos('  ',s),1);
          s:=s+' ';
          while (pos(' ',s)>0) do
            begin
              ss:=copy(s,1,pos(' ',s)-1);
              delete(s,1,pos(' ',s));
              inc(n); len[n]:=length(ss);
            end;
          readln(f1,s);
        end;
    end;
function cr(n:int64):int64; inline;
    begin
      exit(n*n*n);
    end;
procedure solve;
    var
      i,j,line,now:longint;
    begin
      for i:=1 to n do
        sum[i]:=sum[i-1]+len[i];
      for i:=1 to n do
      for j:=1 to n do
        if j<i then cost[i,j]:=oo
        else cost[i,j]:=cr(w-(sum[j]-sum[i-1])-(j-i));

      for i:=1 to n do
        f[0,i]:=oo;
      for i:=1 to l do
        f[i,0]:=cr(w)*i;

      for line:=1 to l do
      for i:=1 to n do
        begin
          f[line,i]:=oo;
          now:=len[i]; j:=i;
          while (now<=w) do
            begin
              f[line,i]:=min(f[line,i],f[line-1,j-1]+cost[j,i]);
              dec(j); if j<1 then break;
              inc(now,len[j]+1);
            end;
        end;
      if f[line,n]<oo then writeln(f2,f[line,n])
      else writeln(f2,-1);
    end;

begin
  openF;
  readln(f1,test);
  for test:=1 to test do
    begin
      inp;
      solve;
    end;
  closeF;
end.

Code mẫu của hieult

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<cassert>
#include<ctime>
#include<algorithm>
#include<iterator>
#include<iostream>
#include<cctype>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<list>
//#include<conio.h>
#define oo 1111111111
#define nmax 100111
#define MAX 111
#define TR(c, it) for(typeof((c).begin()) it=(c).begin(); it!=(c).end(); it++)
double const PI=4*atan(1);

using namespace std;

typedef pair<int, int> II;
typedef vector<int> VI;
typedef vector<II> VII;
typedef vector<VI> VVI;
typedef vector<VII> VVII;

using namespace std;

int test,l,w,f[111][11111],n,a[11111],d;
char s[11111];

int main()
{
    //freopen("PRETTYP.in","r",stdin);
    scanf("%d",&test);
    for(int itest = 1;itest<=test;itest++){
         scanf("%d %d\n",&l,&w);
        // printf("%d %d\n",l,w);
         n = 0;
         memset(a,0,sizeof(a));
         while(gets(s) && strlen(s)!=0){
               for(int i = 0;i<strlen(s);i++){
                     if(s[i]>='a' && s[i]<='z'){
                           if(i==0 || s[i-1]==' ')
                                a[++n] = 1;
                           else a[n]++;
                     }
               }
         }
         //printf("%d\n",n);
         for(int i = 0;i<=l;i++) for(int j = 0;j<=n;j++) f[i][j] = oo;
         f[0][0] = 0;
         for(int i = 1;i<=l;i++){
               for(int j = 1;j<=n;j++){
                    d = a[j];
                    for(int k = j-1;k>=0;k--){
                          if(d>w)
                              break;
                          f[i][j] <?= f[i-1][k] + (w-d)*(w-d)*(w-d);
                          d+=a[k]+1;
                    }
               //printf("%d %d %d\n",i,j,f[i][j]); 
               }
         }
         int kq = oo;
         for(int i = 1;i<=l;i++)
               kq <?= f[i][n]+(l-i)*w*w*w;
         if(kq>=oo) printf("-1\n");
         else printf("%d\n",kq); 
    }
   // getch();
}

Code mẫu của khuc_tuan

#include <iostream>
#include <sstream>
#include <vector>
using namespace std;

int line, w, n;
vector<int> v;
int a[1010];
long long F[110][1010];

int main() {
    int st;
    scanf("%d", &st);
    for(int tt=1;tt<=st;++tt) {
        scanf("%d%d", &line, &w);
        string s;
        getline( cin, s);
        v.resize(0);
        while(getline( cin, s), s!="") {
            istringstream iss(s);
            string s2;
            while(iss>>s2) v.push_back(s2.length());
        }
        n = v.size();
        for(int i=0;i<n;++i)
            a[i] = v[i];
        memset( F, 0x3f, sizeof(F));
        long long inf = F[0][0];
        F[0][0] = 0;
        for(int l=1;l<=line;++l) {
            for(int i=0;i<=n;++i) {
                int cur = 0;
                for(int used=0;;++used) {
                    if(cur > w) break;
                    F[l][i] <?= F[l-1][i-used] + (w-cur) * (w-cur) * (w-cur);
                    if(used==i) break;
                    if(used==0) cur += a[i-1];
                    else cur += 1 + a[i-1-used];
                }
            }
        }
        if(F[line][n] >= inf) cout << -1 << endl;
        else cout << F[line][n] << endl;
    }
    return 0;
}