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

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窮舉基點之後, 對後每一點拉出斜率, 對這些斜率做排序, 求最長(連續相同)的長度即可
效率 O(n*nlogn)

#include <stdio.h>
#include <stdlib.h>
typedef struct {
	int x, y;
}Point;

int cmp(const void *i, const void *j) {
	Point *x, *y;
	x = (Point *)i, y = (Point *)j;
	if(x->x != y->x)
		return x->x - y->x;
	return x->y - y->y;
}
int gcd(int x, int y) {
	if(y == 0)	return x;
	x = abs(x);
	y = abs(y);
	int tmp;
	while(x%y) {
		tmp = x, x = y, y = tmp%y;
	}
	return y;
}
int main() {
	Point D[700];
	int t;
	char str[100];
	scanf("%d", &t);
	getchar();
	getchar();
	while(t--) {
		int n = 0;
		while(gets(str)) {
			if(str[0] == '\0')
				break;
			sscanf(str, "%d %d", &D[n].x, &D[n].y);
			n++;
		}
		qsort(D, n, sizeof(Point), cmp);
		int i, j;
		int ans = 1, g;
		Point slope[700];
		for(i = 0; i < n; i++) {
			int m = 0, tmp;
			for(j = i+1; j < n; j++) {
				slope[m].x = D[i].x - D[j].x;
				slope[m].y = D[i].y - D[j].y;
				g = gcd(slope[m].x, slope[m].y);
				slope[m].x /= g;
				slope[m].y /= g;
				m++;
			}
			qsort(slope, m, sizeof(Point), cmp);
			Point last = slope[0];
			tmp = 1;
			for(j = 1; j < m; j++) {
				if(last.x == slope[j].x && last.y == slope[j].y) {
					tmp++;
				} else {
					if(tmp > ans)
						ans = tmp;
					last = slope[j], tmp = 1;
				}
			}
			if(tmp > ans)
				ans = tmp;
		}
		printf("%d\n", ans+1);
		if(t)	puts("");
	}
    return 0;
}