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

 

1

1 1
2 2
3 3
9 10
10 11

 

Sample Output

 

3

 

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;
}