POJ 3565 - Ants

Description

Young naturalist Bill studies ants in school. His ants feed on plant-louses that live on apple trees. Each ant colony needs its own apple tree to feed itself.

Bill has a map with coordinates of n ant colonies and n apple trees. He knows that ants travel from their colony to their feeding places and back using chemically tagged routes. The routes cannot intersect each other or ants will get confused and get to the wrong colony or tree, thus spurring a war between colonies.

Bill would like to connect each ant colony to a single apple tree so that all n routes are non-intersecting straight lines. In this problem such connection is always possible. Your task is to write a program that finds such connection.

On this picture ant colonies are denoted by empty circles and apple trees are denoted by filled circles. One possible connection is denoted by lines.

Input

The first line of the input file contains a single integer number n (1 ≤ n ≤ 100) — the number of ant colonies and apple trees. It is followed by n lines describing n ant colonies, followed by n lines describing n apple trees. Each ant colony and apple tree is described by a pair of integer coordinates x and y (−10 000 ≤ x, y ≤ 10 000) on a Cartesian plane. All ant colonies and apple trees occupy distinct points on a plane. No three points are on the same line.

Output

Write to the output file n lines with one integer number on each line. The number written on i-th line denotes the number (from 1 to n) of the apple tree that is connected to the i-th ant colony.

Sample Input

5
-42 58
44 86
7 28
99 34
-13 -59
-47 -44
86 74
68 -75
-68 60
99 -60

Sample Output

4
2
1
5
3

分治算法算法思路：
①找到y坐标最小(若有多个，则选择x坐标最小的那个)的点，作为极坐标原点；
②求出其他点的极角的余弦(cos)值，并按照从大到小的顺序排序；
③找到可以与极点相连的对应点(要求确保子问题中有对应相等的点)；
④把连线左边的点和右边的点分解为子问题递归；

#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;


const int N = 100 + 5;
const int C = 1;
const int T = 2;


struct P
{
	int x, y, type, pid;
	double angle;


	void calculate(P& p)
	{
		int dx = x - p.x;
		int dy = y - p.y;
		angle = (double) dx / sqrt((double) (dx * dx + dy * dy));
	}
} p[2 * N];


int ans[N];


int find_basic(const int s, const int t)
{
	int res = -1;
	for(int i = s; i < t; i++)
	{
		if(res == -1 || p[i].y < p[res].y
		|| (p[i].y == p[res].y && p[i].x < p[res].x))
			res = i;
	}


	return res;
}


void calculate(const int s, const int t)
{
	for(int i = s; i < t - 1; i++)
		p[i].calculate(p[t-1]);
}


bool cmp(const P a, const P b)
{
	return a.angle > b.angle;
}


int match(const int s, const int t)
{
	int rec = 0;
	for(int i = s; i < t - 1; i++)
	{
		if(rec == 0 && p[i].type != p[t-1].type)
		{
			if(p[i].type == C)
				ans[p[i].pid] = p[t-1].pid;
			else
				ans[p[t-1].pid] = p[i].pid;
			
			return i;
		}
			
		if(p[i].type == p[t-1].type)
			rec++;
		else	rec--;
	}
}


void solve(const int s, const int t)
{
	int tot = t - s;
	if(tot <= 1)	return;
	
	int b = find_basic(s, t);
	swap(p[b], p[t-1]);
	
	calculate(s, t);
	sort(p + s, p + t - 1, cmp);
	
	int pos = match(s, t);
	solve(s, pos);
	solve(pos + 1, t - 1);
}


int main()
{
	int n;
	cin>>n;
	
	for(int i = 0; i < 2 * n; i++)
	{
		scanf("%d%d", &(p[i].x), &(p[i].y));
		if(i < n)
			p[i].type = C;
		else
			p[i].type = T;
		p[i].pid = (i % n) + 1;	
		//printf("p[%d] = %d\n", i, p[i].pid);
	}
	
	solve(0, 2 * n);
	
	for(int i = 1; i <= n; i++)
		printf("%d\n", ans[i]);


	return 0;
}