来自"NOCOW"

假设我们要让第一个运动员赢得比赛，则必须满足不等式组

L1/v1+L2/u1+L3/w1<L1/v2+L2/u2+L3/w2

L1/v1+L2/u1+L3/w1<L1/v3+L2/u3+L3/w3

L1/v1+L2/u1+L3/w1<L1/v4+L2/u4+L3/w4

…………

L1/v1+L2/u1+L3/w1<L1/vn+L2/un+L3/wn

看L1，L2，L3是否有解即可，将不等式变形

L1/v1+L2/u1+L3/w1<L1/v2+L2/u2+L3/w2

变为

L1/(L3*v1)+L2/(L3*u1)+1/w1<L1/(L3*v2)+L2/(L3*u2)+1/w2

以此类推，那么很显然以L1/L3，L2/L3为元，做半平面交，如果面积大于0则有解，否则无解

program cao;
const
  maxn=200;
  bignum=maxlongint;
  eps=1e-10;
 
type
  Tpoint=record
    x,y:extended;
  end;
 
  Thalf_plane=record
    a,b,c:extended;
  end;
 
  THPI=object  //half plane intersection
    total,tot:longint;
    p,q:array[1..maxn] of Tpoint;
    procedure init;
    function intersect(u,v:Tpoint;tmp:Thalf_plane):Tpoint;
    procedure cut(tmp:Thalf_plane);
    function ok:boolean;
  end;
 
var
  t:THPI;
  a,b,c,d,e,f,g,h,i,j,k,l,n,m,p,q:longint;
  s1,s2,s3:array[0..maxn] of longint;
  tmp:Thalf_plane;
 
procedure THPI.init;
begin
  total:=4;
  p[1].x:=0;
  p[1].y:=0;
  p[2].x:=bignum;
  p[2].y:=0;
  p[3].x:=bignum;
  p[3].y:=bignum;
  p[4].x:=0;
  p[4].y:=bignum;
end;
 
function THPI.intersect(u,v:Tpoint;tmp:Thalf_plane):Tpoint;
var
  ans:Tpoint;
  t1,t2,delta:extended;
begin
  t1:=tmp.a*u.x+tmp.b*u.y+tmp.c;
  t2:=tmp.a*v.x+tmp.b*v.y+tmp.c;
  delta:=-t1/t2;
  ans.x:=(u.x+delta*v.x)/(1+delta);
  ans.y:=(u.y+delta*v.y)/(1+delta);
  exit(ans);
end;
 
procedure THPI.cut(tmp:Thalf_plane);
var
  i,j:longint;
begin
  tot:=0;
  if (abs(tmp.a)<eps)and(abs(tmp.b)<eps)and(tmp.c>=-eps) then
  begin
    total:=0;
    exit;
  end;
  for i:=1 to total do
  begin
    if tmp.a*p[i].x+tmp.b*p[i].y+tmp.c<eps then
    begin
      inc(tot);
      q[tot]:=p[i];
    end
    else
    begin
      j:=i-1;
      if j<=0 then j:=total;
      if tmp.a*p[j].x+tmp.b*p[j].y+tmp.c<-eps then
      begin
        inc(tot);
        q[tot]:=intersect(p[j],p[i],tmp);
      end;
      j:=i+1;
      if j>total then j:=1;
      if tmp.a*p[j].x+tmp.b*p[j].y+tmp.c<-eps then
      begin
        inc(tot);
        q[tot]:=intersect(p[i],p[j],tmp);
      end;
    end;
  end;
  total:=tot;
  move(q[1],p[1],sizeof(Tpoint)*tot);
end;
 
function THPI.ok:boolean;
begin
  exit(total>=3);
end;
 
procedure init;
begin
  read(n);
  for i:=1 to n do
    read(s1[i],s2[i],s3[i]);
end;
 
procedure main;
begin
  for i:=1 to n do
  begin
    t.init;
    for j:=1 to n do
      if i<>j then
      begin
        tmp.a:=1/s1[i]-1/s1[j];
        tmp.b:=1/s2[i]-1/s2[j];
        tmp.c:=1/s3[i]-1/s3[j];
        t.cut(tmp);
        if t.ok=false then break;
      end;
    if t.ok then writeln(‘Yes’) else writeln(‘No’);
  end;
end;
 
begin
  init;
  main;
end.

弱弱的枚举也过了哦

#include <iostream>
 
using namespace std;
 
long n,i,j,k,l,okk,flag,x[101],y[101],z[101],ok[101];
double ct,best;
 
double bbs (double now)
{
       if (now>0) return (now);
                  else return (0-now);
}
 
int main ()
{
    cin>>n;
    for (i=0;i<n;i++)
        cin>>x[i]>>y[i]>>z[i];
    for (j=0;j<=500;j++)
        for (k=0;k<=500-j;k++)
                {
                l=500-j-k;
                best=10000000.00000;
                flag=1;
                for (i=0;i<n;i++)
                    {
                    ct=(double)(j)/(double)(x[i])+(double)(k)/(double)(y[i])+(double)(l)/(double)(z[i]);
                    if (bbs(ct-best)<0.000000000001) {
                                                     flag=0;
                                                     continue;
                                                     }
                    if (ct<best) {
                                 flag=1;
                                 best=ct;
                                 okk=i;
                                 }
                    }
                if (flag) ok[okk]=1;
                }
    for (i=0;i<n;i++)
        if (ok[i]) cout<<"Yes"<<endl;
                   else cout<<"No"<<endl;
    return (0);
}