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
本以为这题用枚举会超时 结果却过了，给了我一个小惊喜。 只是此题要注意的时候 输入的做标可能有负的
[cpp]
#include
#include
#include
#define EQS 1e-7
struct
{
double x,y;
}a[10000];
char s1[10000];
int main()
{
int i,j,n,m,t,l,tag,max,z,s;
double k1,d1,x1,y1,x2,y2,x3,y3;
scanf("%d%*c",&t);
gets(s1);
while(t--)
{
n=0;
while(gets(s1))
{
l=strlen(s1);
if(l==0)
{
break;
}
if(s1[0]=='-')
{
i=1;
}else
{
i=0;
}
for(s=0;i<=l-1;i++)
{
if(s1[i]==' ')
break;
}
s=s*10+s1[i] - '0';
}
if(s1[0]=='-')
{
s=s*-1;
}
a[n].x= (double)s;
if(s1[i+1]=='-')
{
j=i+2;
}else
{
j=i+1;
}
for(s=0;j <= l-1; j++)
{
s=s*10 + s1[j]- '0';
}
if(s1[i+1]=='-')
{
s=s*-1;
}
a[n].y= (double)s;
n++;
}
for(i=0,max=0;i<=n-1; i++)
{
for(j=i+1; j<=n-1;j++)
{
x1 = a[i].x; y1= a[i].y;
x2 = a[j].x; y2= a[j].y;
if(fabs(x2 -x1)<=EQS)
{
tag=0;
d1=x1;
}else
{
tag=1;
k1= (y2 - y1)/(x2 - x1);
d1= y1 - k1 * x1;
}
for(z= j+1,s=2;z<=n-1; z++)
{
if(z==i||z==j)
{
continue;
}
x3= a[z].x; y3= a[z].y;
if(tag==0&&fabs(x3-x1)<=EQS)
{
s++;
}else if(tag==1&&fabs(k1 * x3 +d1- y3)<= EQS)
{
s++;
}
}
if(s>max)
{
max=s;
}
}
}
printf("%d\n",max);
if(t)
{
printf("\n");
}
}
return 0;
}