uva270 - Lining Up(枚举）

给出一些点，看最多有多少个点在同一个直线上。

 

用每一个点作为起点，枚举与其他点的斜率，找出最大值。注意输出格式 和初始化最小ans为2；

 

题目：

 Lining Up 

``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

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The input consists of N pairs of integers, where 1 < N < 700. Each pair of integers is separated by one blank and ended by a new-line character. The list of pairs is ended with an end-of-file character. No pair will occur twice.

 

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
The output consists of one integer representing the largest number of points that all lie on one line.

 

Sample Input

1

1 1
2 2
3 3
9 10
10 11

 

Sample Output

3


代码：

#include <iostream>
#include <string>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
double K[1000000];
double X[1000];
double Y[1000];
int main()
{
    freopen("in.txt","r",stdin);
    int n;
    char tmp[10];
    cin>>n;
    cin.get();
    cin.get();
    while(n--)
    {
        int p=0;
       // getchar();
        while(gets(tmp))
        {
            //cout<<"tmp"<<tmp<<endl;
            if(!tmp[0])break;
            sscanf(tmp,"%lf %lf",&X[p],&Y[p]);
           // cout<<"X Y "<<X[p]<<" "<<Y[p]<<endl;
            p++;
        }
        int pk=0,ans=2;
        for(int i=0;i<p;i++)
        {
            pk=0;
            for(int j=0;j<p;j++)
            {
                if(i==j)continue;
                //cout<<Y[j]<<" "<<Y[i]<<" "<<X[j]<<" "<<X[i]<<endl;
                K[pk++] = (Y[j]-Y[i])/(X[j]-X[i]);
                //cout<<K[pk-1]<<endl;
            }
            sort(K,K+pk);
        //cout<<"ok"<<endl;
            int tans=2;
            for(int i=0;i<pk-1;i++)
            {
                int j=i,m=j+1;
                tans=2;
                while(K[j]==K[m])
                {
                    tans++;j++;m++;
                }
                if(tans>ans)ans=tans;
            }
        }
        cout<<ans<<endl;
        if(n)cout<<endl;
    }
    return 0;
}