[POJ1118]Lining Up

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 24071		Accepted: 7564

Description

"How am I ever going to solve this problem?" said the pilot.

Indeed, the pilot was not facing an easy task. She had to drop packages at specific points scattered in a dangerous area. Furthermore, the pilot could only fly over the area once in a straight line, and she had to fly over as many points as possible. All points were given by means of integer coordinates in a two-dimensional space. The pilot wanted to know the largest number of points from the given set that all lie on one line. Can you write a program that calculates this number?

Your program has to be efficient!

Input

Input consist several case,First line of the each case is an integer N ( 1 < N < 700 ),then follow N pairs of integers. Each pair of integers is separated by one blank and ended by a new-line character. The input ended by N=0.

Output

output one integer for each input case ,representing the largest number of points that all lie on one line.

Sample Input

5
1 1
2 2
3 3
9 10
10 11
0

Sample Output

3

Source

East Central North America 1994

THINKING

　　搜索，分别判断X轴，Y轴，斜率即可。

type point=record
x,y:longint;
end;

var a:array[0..5000] of point;
f:array[0..5000] of longint;
i,j,ans,sum,pred,n,xx,yy,g,k,aa,bb,x:longint;

function max(x,y:longint):longint;
begin
if x>y then exit(x) else exit(y);
end;

procedure sort(l,r: longint);
var
i,j,x: longint;y:point;
begin
i:=l;
j:=r;
x:=a[(l+r) div 2].x;
repeat
while a[i].x<x do
inc(i);
while x<a[j].x do
dec(j);
if not(i>j) then
begin
y:=a[i];
a[i]:=a[j];
a[j]:=y;
inc(i);
j:=j-1;
end;
until i>j;
if l<j then
sort(l,j);
if i<r then
sort(i,r);
end;

procedure sort1(l,r: longint);
var
i,j,x: longint;y:point;
begin
i:=l;
j:=r;
x:=a[(l+r) div 2].y;
repeat
while a[i].y<x do
inc(i);
while x<a[j].y do
dec(j);
if not(i>j) then
begin
y:=a[i];
a[i]:=a[j];
a[j]:=y;
inc(i);
j:=j-1;
end;
until i>j;
if l<j then
sort1(l,j);
if i<r then
sort1(i,r);
end;

procedure main;
begin
if n<=2 then
begin
writeln(n);
halt;
end;
if n=3 then
if(abs(a[2].x-a[1].x)=abs(a[3].x-a[2].x))and(abs(a[2].y-a[1].y)=abs(a[3].y-a[2].y)) then
begin
writeln(3);
halt;
end
else
begin
writeln(2);
halt;
end;
//特殊判断
sort(1,n);
ans:=0;
sum:=1;
pred:=a[1].x;
for i:=2 to n do
if a[i].x=pred then inc(sum)
else
begin
ans:=max(ans,sum);
pred:=a[i].x;
sum:=1;
end;
//判断横
ans:=max(ans,sum);
sort1(1,n);
sum:=1;
pred:=a[1].y;
for i:=2 to n do
if a[i].y=pred then inc(sum)
else
begin
ans:=max(ans,sum);
pred:=a[i].y;
sum:=1;
end;
//判断纵
ans:=max(ans,sum);
for i:=1 to n do
for j:=i+1 to n do
begin
sum:=2;
for k:=j+1 to n do
begin
aa:=(a[i].y-a[k].y)*(a[j].x-a[k].x);
bb:=(a[i].x-a[k].x)*(a[j].y-a[k].y);
if aa=bb then inc(sum);
end;
ans:=max(ans,sum);
end;
//判断斜率
writeln(ans);
end;

begin
while true do
begin
readln(n);
if n=0 then halt;
for i:=1 to n do
readln(a[i].x,a[i].y);
main;
end;
end.

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