POJ2780 Linearity【水题】

Linearity Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 8517 Accepted: 1969

Description

Alice often examines star maps where stars are represented by points in a plane and there is a Cartesian coordinate for each star. Let the Linearity of a star map be the maximum number of stars in a straight line. For example, look at the star map shown on the figure above, the Linearity of this map is 3, because the star 1, star 2, and star 5 are within the same straight line, and there is no straight line that passes 4 stars. You are to write a program to find the Linearity of a star map.

Input

Input will contain multiple test cases. Each describes a star map. For each test case, the first line of the input contains the number of stars N (2 <= N <= 1000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0 <= X, Y <= 1000). There can be only one star at one point of the plane.

Output

Output the Linearity of the map in a single line.

Sample Input

5 0 0 2 0 0 2 1 1 2 2

Sample Output

3

Source

POJ Monthly--2006.03.26,static

问题链接：POJ2780 Linearity。

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：

  程序中，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。

  这个问题与POJ1118 HDU1432 Lining Up基本上相同，只是输入数据格式略有不同。

AC的C语言程序如下：

/* POJ2780 Linearity */ #include <stdio.h> #define MAXN 1000 struct { int x, y; } p[MAXN+1]; /* point */ int main(void) { int n, ans, max, i, j, k; while(scanf("%d", &n) != EOF) { 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; }