UVa 270/poj 1118 lining up: Computational geometry

270-lining up

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_ problem&problem=206

http://poj.org/problem?id=1118

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

Indeed, the pilot is not facing a 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 Pos Sible. All points were given by means of the integer coordinates in a two-dimensional spaces. The pilot wanted to know the largest number of points from the given set, all lie on one line. Can you write a program so calculates this number?

Your program has to be efficient!

Input

The input begins with a single positive integer in a line by itself indicating the number of cases following, each of them as described below. This are followed by a blank line, and there are also a blank line between two consecutive. The input consists of N pairs of integers, where 1 < N < 700. Each pair of integers was 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 would occur twice.

Output

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

Sample Input

1

1 1
2 2 3 3 9
ten
10 11

Sample Output

3

Ideas:

At any point, calculate the other points to the point of the slope after the order, the slope of the same points must be in a straight line. Complexity: O (N^2*log N)

UVA Code:

/*0.129s*/#include <cstdio> #include <cmath> #include <algorithm> using namespace std; const INT
  
INF = 1 << 30;
Char str[20];
  
Double x[705], y[705], d[705];
    int main (void) {int T, n, I, J, K;
    int ans, count; scanf ("%d\n", &t);///read a newline ~ while (t--) {for (n = 0; gets (str); ++n) {if (str[0
            ] = = "") break;
        SSCANF (str, "%LF%LF", &x[n], &y[n]);
        //////ans = 1; for (i = 0; i < n-1. ++i) {for (j = i + 1, k = 0; J < N; ++j, ++k) d[k] = (x[ J] = = X[i]?
            INF: (Y[j]-y[i])/(X[j]-x[i));
            Sort (d, D + K);
            Count = 1;
                    for (j = 1; j < K; ++j) {if (Fabs (D[J)-d[j-1]) < 1e-9) {
                    ++count;
                ans = max (ans, count);
            else count = 1;
 }
        }       printf ("%d\n", ans + 1);
    if (t) putchar (' \ n ');
return 0; }

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

Code for POJ:

/*125ms,204kb*/
  
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace Std;
const int INF = 1 <<;
  
Char str[20];
Double x[705], y[705], d[705];
  
int main (void)
{
    int n, I, J, K;
    int ans, count;
    while (scanf ("%d", &n), N)
    {for
        (i = 0; i < n; ++i)
            scanf ("%lf%lf", &x[i), &y[i]);
        ans = 1;
        for (i = 0; i < n-1. ++i)
        {
            for (j = i + 1, k = 0; J < N; ++j, ++k)
                d[k] = (x[j] = = X[i]? INF: (Y[j]-y[i])/(X[j]-x[i));
            Sort (d, D + k);
            Count = 1;
            for (j = 1; j < K; ++j)
            {
                if (fabs (D[J)-d[j-1]) < 1e-9)
                {
                    ++count;
                    ans = max (ans, count);
                else count = 1;
            }
        }
        printf ("%d\n", ans + 1);
    }
    return 0;
}