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

Code mẫu của ladpro98

#include <bits/stdc++.h>
#define sqr(x) ((x)*(x))
const int N = 1111;
using namespace std;
struct P {
    int x; int y;
};
P a[N], b[N];
double height(P a, P b, P c) {
    double S = abs(a.x*(b.y-c.y)+b.x*(c.y-a.y)+c.x*(a.y-b.y));
    double AC2 = sqr(a.x-c.x)+sqr(a.y-c.y);
    double AB2 = sqr(a.x-b.x)+sqr(a.y-b.y);
    double BC2 = sqr(b.x-c.x)+sqr(b.y-c.y);
    if (AB2 > AC2 + BC2) return sqrt(AC2);
    if (AC2 > AB2 + BC2) return sqrt(AB2);
    return S / (sqrt(BC2));
}

int main()
{
    //freopen("LEM.in", "r", stdin);
    int n, m, i, j; double res = 1e9;
    scanf("%d %d\n", &n, &m);
    for(i=1; i<=n; i++) scanf("%d %d\n", &a[i].x, &a[i].y);
    for(i=1; i<=m; i++) scanf("%d %d\n", &b[i].x, &b[i].y);
    for(i=1; i<=n; i++)
        for(j=1; j<m; j++)
            res = min(res, height(a[i], b[j], b[j+1]));
    for(j=1; j<=m; j++)
        for(i=1; i<n; i++)
            res = min(res, height(b[j], a[i], a[i+1]));
    printf("%.3f", res);
    return 0;
}

Code mẫu của RR

{$R+,Q+,N+}
uses math;
const
  FINP='';
  FOUT='';
  MAXN=1001;
  oo=1000000001;
var
  f1,f2:text;
  x1,y1,x2,y2:array[1..MAXN] of double;
  m,n:longint;
  kq:double;
procedure openF;
begin
  assign(f1,FINP); reset(f1);
  assign(f2,FOUT); rewrite(f2);
end;
procedure closeF;
begin
  close(f1); close(f2);
end;
procedure inp;
begin
  read(f1,n,m);
  for n:=1 to n do read(f1,x1[n],y1[n]);
  for m:=1 to m do read(f1,x2[m],y2[m]);
end;
procedure ans;
begin
  writeln(f2,kq:0:3);
end;
function kc(x1,y1,x2,y2:double):double;
begin
  kc:=sqrt(sqr(x1-x2)+sqr(y1-y2));
end;
function dist(x,y,x1,y1,x2,y2:double):double;
var
  a,b,t,xh,yh:double;
begin
  a:=x2-x1;
  b:=y2-y1;
  t:=(a*(x-x1)+b*(y-y1))/(a*a+b*b);
  xh:=x1+a*t; yh:=y1+b*t;
  if (yh>max(y1,y2)) or (yh<min(y1,y2)) then exit(oo)
  else exit(kc(x,y,xh,yh));
end;
procedure solve;
var
  i,j:longint;
begin
  kq:=oo;
  for i:=1 to n do
  for j:=1 to m do
    kq:=min(kq,kc(x1[i],y1[i],x2[j],y2[j]));
  for i:=1 to n-1 do
  for j:=1 to m-1 do
    begin
      kq:=min(kq,dist(x1[i],y1[i],x2[j],y2[j],x2[j+1],y2[j+1]));
      kq:=min(kq,dist(x2[j],y2[j],x1[i],y1[i],x1[i+1],y1[i+1]));
    end;
end;
begin
  openF;
  inp;
  solve;
  ans;
  closeF;
end.