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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
11 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
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