HDU5020（判断三点共线）BestCoder Round #10（1003）

最新推荐文章于 2021-01-11 15:34:30 发布

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#ACM #hdu #bestcoder

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本文主要探讨了在BestCoder Round #10（1003）比赛中遇到的HDU5020题目，该题涉及到几何中的三点共线判断问题。通过对几何原理的深入分析，文章揭示了解决此类问题的关键思路和算法实现。

Revenge of Collinearity

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 184 Accepted Submission(s): 51

Problem Description

In geometry, collinearity is a property of a set of points, specifically, the property of lying on a single line. A set of points with this property is said to be collinear (often misspelled as colinear).
---Wikipedia

Today, Collinearity takes revenge on you. Given a set of N points in two-dimensional coordinate system, you have to find how many set of <P_i, P_j, P_k> from these N points are collinear. Note that <P_i, P_j, P_k> cannot contains same point, and <P_i, P_j, P_k> and <P_i, P_k, P_j> are considered as the same set, i.e. the order in the set doesn’t matter.

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case begins with an integer N, following N lines, each line contains two integers Xi and Yi, describing a point.

[Technical Specification]
1. 1 <= T <= 33
2. 3 <= N <= 1 000
3. -1 000 000 000 <= Xi, Yi <= 1 000 000 000, and no two points are identical.
4. The ratio of test cases with N > 100 is less than 25%.

Output

For each query, output the number of three points set which are collinear.

Sample Input

Sample Output

1
0

题意：给出n个点，找出三点贡献的点的集合

思路：n=1000，我比赛的时候实在没思路，虽然知道暴力n^3一定会超时，但还也写完意思的交了一发，平时几乎不碰几何党的飘过~

赛后看了某神犇的代码，思路瞬间开阔了

就是将n个点按x，y从小到大排序，x为第一关键字，y为第二关键字

排完序以后就是两层for暴搜了，因为已经排好序，就可以按斜率将所有斜率一样的点归为一个集合

假设当前要判断的点为i，有p个点与之共线，因为i是一定要选的，所以只需在剩下的p个点中选择2个就好了，为p*(p-1)/2

累加这些就是答案了

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

struct node{
    int x,y;
}vex[1010];

typedef pair pii;
map< pii , int >mp;
map< pii , int >::iterator it;

int gcd(int a,int b)
{
    if(a==0)return b;
    if(b==0)return a;
    if(a>b)swap(a,b);
    while(b)
    {
        int temp=a%b;
        a=b;
        b=temp;
    }
    return a;
}

bool cmp(node a,node b)
{
    if(a.x==b.x)return a.ysecond >=2 ){
                    int x=it->second;
                    ans+=x*(x-1)/2;
                }
            }
        }

        printf("%d\n",ans);

    }
    return 0;
}