# POJ 3668 Game of Lines C语言版

POJ题解 2737浏览

Description

Farmer John has challenged Bessie to the following game: FJ has a board with dots marked at N (2 ≤ N ≤ 200) distinct lattice points. Dot i has the integer coordinates Xi and Yi (-1,000 ≤ Xi ≤ 1,000; -1,000 ≤ Yi ≤ 1,000).

Bessie can score a point in the game by picking two of the dots and drawing a straight line between them; however, she is not allowed to draw a line if she has already drawn another line that is parallel to that line. Bessie would like to know her chances of winning, so she has asked you to help find the maximum score she can obtain.

Input

* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 describes lattice point i with two space-separated integers: Xi and Yi.

Output

* Line 1: A single integer representing the maximal number of lines Bessie can draw, no two of which are parallel.

Sample Input

4
-1 1
-2 0
0 0
1 1

Sample Output

4

Slyar:题目大意就是求N个点最多确定多少条互不平行的直线。

```#include
#include
#include

double k[40005];

/* 注意双精度比较函数不能写成减法 */
int cmp(double* a, double* b)
{
if (*a < *b) return -1;
return 1;
}

int main()
{
double x[201], y[201];
double dx, dy;
int i, j, n, ans, sum, sign;
scanf("%d", &n);
for (i = 0; i < n; i++)
{
scanf("%lf%lf", &x[i], &y[i]);
}
sign = 0;
sum = 0;
ans = 1;
for (i = 0; i < n; i++)
{
for (j = i + 1; j < n; j++)
{
dy = y[i] - y[j];
dx = x[i] - x[j];
/* 双精度不能直接与0判等 */
if (fabs(dx) < 1e-6)
{
if (sign == 0) ans++;
sign = 1;
}
else
{
k[sum++] = dy / dx;
}
}
}
qsort(k, sum, sizeof(double), cmp);
for (i = 1; i < sum; i++)
{
/* 因为精度问题也不能写成减法 */
if (k[i] > k[i-1]) ans++;
}
printf("%d\n", ans);
//system("pause");
return 0;
}
```

### 网友最新评论 (2)

1. 不能和0比较，可以和0.0比较啊！
匿名10年前 (2010-10-13)回复
2. 这一个礼拜我都不用来你博客了……全是鸟语……
Jutoy12年前 (2009-04-21)回复