Aizu —— 1379 Parallel Lines（暴力搜索）

Problem B

Given an even number of distinct planar points, consider coupling all of the points into pairs. All the possible couplings are to be considered as long as all the given points are coupled to one and only one other point.

When lines are drawn connecting the two points of all the coupled point pairs, some of the drawn lines can be parallel to some others. Your task is to find the maximum number of parallel line pairs considering all the possible couplings of the points.

For the case given in the first sample input with four points, there are three patterns of point couplings as shown in Figure B.1. The numbers of parallel line pairs are 0, 0, and 1, from the left. So the maximum is 1.

Figure B.1. All three possible couplings for Sample Input 1

For the case given in the second sample input with eight points, the points can be coupled as shown in Figure B.2. With such a point pairing, all four lines are parallel to one another. In other words, the six line pairs (L1,L2)(L1,L2), (L1,L3)(L1,L3), (L1,L4)(L1,L4), (L2,L3)(L2,L3), (L2,L4)(L2,L4) and (L3,L4)(L3,L4) are parallel. So the maximum number of parallel line pairs, in this case, is 6.

Input

The input consists of a single test case of the following format.

mm
x1x1 y1y1
...
xmxm ymym

Figure B.2. Maximizing the number of parallel line pairs for Sample Input 2

The first line contains an even integer mm, which is the number of points (2≤m≤162≤m≤16). Each of the following mm lines gives the coordinates of a point. Integers xixi and yiyi (−1000≤xi≤1000,−1000≤yi≤1000−1000≤xi≤1000,−1000≤yi≤1000) in the ii-th line of them give the xx- and yy-coordinates, respectively, of the ii-th point.

The positions of points are all different, that is, xi≠xjxi≠xj or yi≠yjyi≠yj holds for all i≠ji≠j. Furthermore, No three points lie on a single line.

Output

Output the maximum possible number of parallel line pairs stated above, in one line.

Sample Input 1

Sample Output 1

Sample Input 2

Sample Output 2

Sample Input 3

Sample Output 3

Sample Input 4

2
-1000 1000
1000 -1000

Sample Output 4

Sample Input 5

Sample Output 5

这个题当时想暴力，没敢写，主要是多种斜率存在情况下，不会写了，原来是搜索啊，果然搜索还是很牛逼！！

题意：坐标轴中，给出偶数个点，任意两点相连，问互相平行的线最多有多少。

题解：不一定是一种斜率就是最多，可以有多种不同~~~坑~~

上代码：

#include <iostream>
using namespace std;
const int MAX = 21;
struct Point{
	int x,y;
}point[MAX];//存点坐标
struct Vector{
	int a,b;
}vector[MAX];//存直线向量
int n,maxx,cnt;
bool vis[MAX];
void jisuan(){
	int ans=0;
	for (int i = 0; i < cnt;i++){
		for (int j = i+1; j < cnt;j++){
			if(vector[i].a*vector[j].b==vector[i].b*vector[j].a) ans++;//向量判断是否平行
		}
	}
	maxx=max(maxx,ans);
	return;
}
void dfs(int index){
	while(vis[index]) index++;//找过的点不找了，很重要，不然会T而且会错
	if(index>=n){
		jisuan();
		return;
	}
	vis[index]=true;
	for (int i =  0; i < n;i++){
		if(!vis[i]&&index!=i){//没用过且不是同一个点
			vis[i]=1;
			int xx=point[i].x-point[index].x;//计算直线向量
			int yy=point[i].y-point[index].y;
			vector[cnt].a=xx;
			vector[cnt++].b=yy;
			dfs(index+1);
			cnt--;
			vis[i]=false;
		}
	}
	vis[index]=0;//这个别忘了回溯
	return;
}
int main(){
	cin >> n;
	for (int i = 0;i < n;i++){
		cin >> point[i].x >> point[i].y;
	}
	maxx=0;
	dfs(0);
	cout << maxx << endl;
	return 0;
}

果然水题，只是想不到怎样搜~~~加强训练啊~~