【POJ 3668】Game of Lines

Game of Lines

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6555		Accepted: 2439

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 X_i and Y_i (-1,000 ≤ X_i ≤ 1,000; -1,000 ≤ Y_i ≤ 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: X_i and Y_i.

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

Source

USACO 2008 February Silver

 

然而此题只是简单的枚举。

枚举一个点和另一个点匹配，并且记录这条直线的斜率，如果发现了斜率曾经出现过就忽略。

注：由于直线可能与y轴平行（即x[i] == x[j]）所以此时设为无穷大（然而只是2³¹ - 1）。

#include<cstdio>
#include<cstring>
#include<set>
using namespace std;

int x[205];
int y[205];
int n;
set<double> s;

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; ++i)
        scanf("%d%d", &x[i], &y[i]);
    int ans = 0;
    s.clear();
    for (int i = 0; i < n; ++i)
        for (int j = i + 1; j < n; ++j)
        {
            double k;
            if (x[i] - x[j] == 0) k = 0x7fffffff;
            else k = (double)(y[i] - y[j]) / (x[i] - x[j]);
            if (!s.count(k))
            {
                ++ans;
                s.insert(k);
            }
        }
    printf("%d
", ans);
    return 0;
}