CodeForces 196C.Paint Tree(分治+极角排序)

C. Paint Tree

time limit per test 2 seconds
memory limit per test 256 megabytes

You are given a tree with $n$ vertexes and $n$ points on a plane, no three points lie on one straight line.

Your task is to paint the given tree on a plane, using the given points as vertexes.

That is, you should correspond each vertex of the tree to exactly one point and each point should correspond to a vertex. If two vertexes of the tree are connected by an edge, then the corresponding points should have a segment painted between them. The segments that correspond to non-adjacent edges, should not have common points. The segments that correspond to adjacent edges should have exactly one common point.

Input
The first line contains an integer $n(1\le n\le 1500)$ — the number of vertexes on a tree (as well as the number of chosen points on the plane).

Each of the next $n-1$ lines contains two space-separated integers $u_i $and $v_i(1\le u_i,v_i\le n,u_i̸=v_i)$ — the numbers of tree vertexes connected by the $i$-th edge.

Each of the next $n$ lines contain two space-separated integers $x_i $and$ y_i ( - 10^9 ≤ x_i, y_i ≤ 10^9)$ — the coordinates of the $i$-th point on the plane. No three points lie on one straight line.

It is guaranteed that under given constraints problem has a solution.

Output
Print $n$ distinct space-separated integers from $1$ to $n$: the $i$-th number must equal the number of the vertex to place at the $i$-th point (the points are numbered in the order, in which they are listed in the input).

If there are several solutions, print any of them.

Examples
input1

3
1 3
2 3
0 0
1 1
2 0

output1

1 3 2

input2

4
1 2
2 3
1 4
-1 -2
3 5
-3 3
2 0

output2

4 2 1 3

Note
The possible solutions for the sample are given below.

Solution

对于每一棵树，以最左下角的点作为该树的根，进行极角排序，再根据该树每棵子树的大小分配点的数量，进行递归。由于不存在三点共线所以解是一定存在的。
如图所示：

Code

#include <cstdio>
#include <algorithm>
#define N 1510
using namespace std;
typedef long long LL;

struct Node {
	LL to, nxt;
}e[N << 1];

struct Point {
	LL x, y, id;
}p[N], pp[N], poi;

LL cnt, n, lst[N], ans[N], size[N];

bool cmp(Point a, Point b) {//按极角排序
	LL ax = a.x - poi.x, ay = a.y - poi.y, bx = b.x - poi.x, by = b.y - poi.y;
	if (ax >= 0 && bx <= 0) return 1;
	if (ax <= 0 && bx >= 0) return 0;
	return (ax * by > bx * ay);
}

inline void add(LL u, LL v) {
	e[++cnt].to = v;
	e[cnt].nxt = lst[u];
	lst[u] = cnt;
}

void predfs(LL x, LL fa) {//预处理出子树大小
	size[x] = 1;
	for (int i = lst[x]; i; i = e[i].nxt) {
		if (e[i].to == fa) continue;
		predfs(e[i].to, x);
		size[x] += size[e[i].to];
	}
}

void dfs(LL l, LL r, LL x, LL fa)//对区间进行分治
	LL mini = l;
	for (int i = l + 1; i <= r; ++i)
		if (p[i].y < p[mini].y || p[i].y == p[mini].y && p[i].x < p[mini].x) mini = i;
	swap(p[l], p[mini]);
	ans[p[l].id] = x;
	poi = p[l];
	sort(p + l + 1, p + r + 1, cmp);
	LL now = l + 1;
	for (LL i = lst[x]; i; i = e[i].nxt) {
		if (e[i].to == fa) continue;
		dfs(now, now + size[e[i].to] - 1, e[i].to, x);
		now += size[e[i].to];
	}
}

int main() {
	scanf("%lld", &n);
	for (LL i = 1; i < n; ++i) {
		LL a, b;
		scanf("%lld%lld", &a, &b);
		add(a, b);
		add(b, a);
	}
	for (LL i = 1; i <= n; ++i) {
		scanf("%lld%lld", &p[i].x, &p[i].y);
		p[i].id = i;
	}
	predfs(1, 1);
	dfs(1, n, 1, 1);
	for (LL i = 1; i < n; ++i) {
		printf("%lld ", ans[i]);
	}
	printf("%lld\n", ans[n]);
	return 0;
}