POJ1118 HDU1432 Lining Up【水题】

Lining Up Time Limit: 2000MS Memory Limit: 32768KTotal Submissions: 26016 Accepted: 8153

Description

"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

Input consist several case,First line of the each case is an integer N ( 1 < N < 700 ),then follow N pairs of integers. Each pair of integers is separated by one blank and ended by a new-line character. The input ended by N=0.

Output

output one integer for each input case ,representing the largest number of points that all lie on one line.

Sample Input

5 1 1 2 2 3 3 9 10 10 11 0

Sample Output

3

Source

East Central North America 1994

问题链接：POJ1118 HDU1432 Lining Up。

题意简述：输入n，输入n个整数对，即n个坐标点，问最多共线点数是多少。

问题分析：用暴力法解决这个问题，好在计算规模不算大。

程序说明：判断共线时，使用的是乘法，没有用除法，可以保证精确的计算结果。

AC的C语言程序如下：

/* POJ1118 HDU1432 Lining Up */ #include <stdio.h> #define MAXN 700 struct { int x, y; } p[MAXN]; /* point */ int main(void) { int n, ans, max, i, j, k; while(scanf("%d", &n) != EOF && n != 0) { for(i=0; i<n; i++) scanf("%d%d", &p[i].x, &p[i].y); ans = 2; for(i=0; i<n; i++) for(j=i+1; j<n; j++) { max = 2; for(k=j+1; k<n; k++) if ((p[j].x - p[i].x)*(p[k].y - p[j].y) == (p[j].y - p[i].y)*(p[k].x - p[j].x)) max++; if(max > ans) ans = max; } printf("%d\n", ans); } return 0; }