[POJ1118]Lining Up | 易学教程

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

Sample Output

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.

View Code

来源：https://www.cnblogs.com/yangqingli/p/4889105.html

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