UVA 270 Lining Up

题意：给你n个点，求出能在一条直线上的点的最大值。

 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

11 12 23 39 1010 11

Sample Output

3

 

#include <iostream>#include <cstdio>#include<cstring>#include <cmath>using namespace std;int x[10005],y[10005];int cnt;int tmp,Max;void found()  {      Max=1;     int i,j,k;    for(i=0;i<cnt;i++)      for(j=i+1;j<cnt;j++)    {        tmp=2;       for(k=j+1;k<cnt;k++)         {             if((y[j]-y[i])*(x[k]-x[j])==(y[k]-y[j])*(x[j]-x[i]))             tmp++;         }    Max=max(Max,tmp);    }    printf("%d\n",Max);  }int main(){    int T;    char str[110];   scanf("%d%*c",&T);     getchar();   while(T--)   {       cnt=0;     memset(x,0,sizeof(x));     memset(y,0,sizeof(y));       while(gets(str))       {           if(!str[0]&&cnt) break;           sscanf(str,"%d%d",&x[cnt],&y[cnt]);           cnt++;       }      found();    if(T) printf("\n");   }    return 0;}