UVa 270 / POJ 1118 Lining Up (计算几何)

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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 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



思路:

任取一点,计算出其他点到该点斜率后排序,斜率相同的点必然在一条直线上。 
复杂度:O(N^2*log N) 


UVa的代码:

/*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);///多读一个换行~
	while (t--)
	{
		for (n = 0; gets(str); ++n)
		{
			if (str[0] == '\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;
}


POJ的代码:

/*125ms,204KB*/

#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 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;
}


Source


East Central North America 1994

阿里云国内75折 回扣 微信号:monov8
阿里云国际,腾讯云国际,低至75折。AWS 93折 免费开户实名账号 代冲值 优惠多多 微信号:monov8 飞机:@monov6

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