UVA 270 Lining Up (几何判断共线点)

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

Sample Output

题意：给定一些点坐标。求共线点最多的个数。。

思路：我的做法是暴力枚举。每次先枚举出2个点。作为直线。再去枚举第三个点看在不在直线上。可以推出一个公式

y(x1 - x2) = (y1 - y2) * x + y2 * x1 - y1 * x2的时候。为(x,y)在(x1,y1)和(x2,y2)组成的直线上。。不过这样做的话时间复杂度为O(n^3)..中间有个优化。就是如果两个点之前判断连接过了。之后在遇到就直接跳过。。但是依然跑了快2秒。- -

看网上别人做法有一种时间复杂度为O(n^2logn)的做法。。是每次枚举一个点作为原点。然后把其他点和它的斜率算出来。然后找出这些斜率中相同斜率出现次数最多的作为最大值。感觉不错。

我的代码：

#include <stdio.h>

#include <string.h>

int t;

int n, i, j, k, l;

int max, ans;

int vis[705][705];

int mark[705];

char sb[30];

struct Point {

	int x, y;

} p[705];

int main() {

	scanf("%d%*c%*c", &t);

	while (t --) {

		n = 0; max = 0;

		memset(vis, 0, sizeof(vis));

		while (gets(sb) && sb[0] != '\0') {

			sscanf(sb, "%d%d", &p[n].x, &p[n].y);

			n ++;

		}

		for (i = 0; i < n; i ++)

			for (j = i + 1; j < n; j ++) {

				if (vis[i][j]) continue;

				ans = 0;

				for (k = 0; k < n; k ++) {

					if (p[k].y * (p[i].x - p[j].x) == (p[i].y - p[j].y) * p[k].x + p[j].y * p[i].x - p[i].y * p[j].x) {

						mark[ans ++] = k;

						for (l = 0; l < ans - 1; l ++)

							vis[k][mark[l]] = vis[mark[l]][k] = 1;

					}

				}

				if (max < ans)

					max = ans;

			}

		printf("%d\n", max);

		if (t) printf("\n");

	}

	return 0;

}