POJ3668 Game of Lines

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6791		Accepted: 2523

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

 

 

计算每条线的斜率，然后统计不同斜率的个数就可以。

又陷入了一WA就是好几次的窘境啊……

先是RE，反映半天原来是数组开小了，然后各种WA，于是看了评论区，发现要调精度到1e-8

……

/**/
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const double eps=1e-8;
const int mxn=3000;
double x[mxn],y[mxn];
int n;
double k[mxn*20];
int cnt;
int main(){
    scanf("%d",&n);
    int i,j;
    for(i=1;i<=n;i++)scanf("%lf%lf",&x[i],&y[i]);
    for(i=1;i<n;i++)
      for(j=i+1;j<=n;j++){
            if(x[i]==x[j])k[++cnt]=1e15;
            else k[++cnt]=(double)(y[i]-y[j])/(x[i]-x[j]);
      }
    sort(k+1,k+cnt+1);
    int ans=0;
        ans++;//一定可以画至少一条
    for(i=1;i<cnt;i++){
        if(fabs(k[i]-k[i+1])>eps)ans++;
    }
    cout<<ans;
    return 0;
}