Revenge of Collinearity
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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 <Pi, Pj, Pk> from these N points are collinear. Note that <Pi, Pj,
Pk> cannot contains same point, and <Pi, Pj, Pk> and <Pi, Pk, Pj> 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
2 3 1 1 2 2 3 3 4 0 0 1 0 0 1 1 1
Sample Output
1 0
Source
题目大意：给出n个二维平面的点，问有多少对三点共线。
解题思路：既然是三点共线，不妨设为A，B，C三点，如果我们纯粹暴力枚举A，B，C三点，为了ABC三点之间不相互重复，我们可以先对点排个序，按照x优先非递减，y次之非递减。但是纯粹暴力必然TLE，如果减少枚举的点数，那么就可以减少时间复杂度，假如我们只枚举点A，那么对于其他点中的每一个都会与A点形成一条直线，如果某一条直线出现的不止一次，那么就意味着至少有另外两对与A点在同一直线上，那么就相应的存在三点共线。假设有k个点与A点共线（注意：这里共线是说这k+1个点都在同一直线上）（另外，还要注意这k个点是不重复的，何为不重复呢？举例说明：有(1,0)，(2,0)，(3,0)，(4,0)，(5,0)五个点，当A=(1,0)时，k=4；当A=(2,0)时，k=3而不是k=4；当A=(3,0)时，k=2，……）。因为枚举的是A点，其中A点必包括，那么就有k*(k-1)/2对三点共线，枚举完所有的A点，总和就是答案。
对于共线数目k的求解………………………………
代码如下：
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
#include <limits.h>
#define debug "output for debug\n"
#define pi (acos(-1.0))
#define eps (1e-6)
#define inf (1<<28)
#define sqr(x) (x) * (x)
#define mod 1000000007
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
typedef pair<ll,ll> p;
map<p,ll> m;
struct point
{
ll x,y;
}a[1005];
ll gcd(ll a,ll b)
{
return b?gcd(b,a%b):a;
}
bool cmp(point a,point b)
{
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
int main()
{
ll i,j,k,n,t;
scanf("%I64d",&t);
while(t--)
{
scanf("%I64d",&n);
for(i=0;i<n;i++)
scanf("%I64d%I64d",&a[i].x,&a[i].y);
sort(a,a+n,cmp);
ll ans=0;
for(i=0;i<n;i++)
{
m.clear();
for(j=i+1;j<n;j++)
{
ll dx=a[j].x-a[i].x;
ll dy=a[j].y-a[i].y;
ll d=gcd(dx,dy);
dx=dx/d;
dy=dy/d;
m[make_pair(dx,dy)]++;
}
map<p,ll>::iterator it;
for(it=m.begin();it!=m.end();it++)
{
k=it->second;
if(k>=2)
ans=ans+k*(k-1)/2;
}
}
printf("%I64d\n",ans);
}
return 0;
}
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