题目链接 Paint Tree

给你一棵n个点的树和n个直角坐标系上的点,现在要把树上的n个点映射到直角坐标系的n个点中,要求是除了在顶点处不能有线段的相交。

我们先选一个在直角坐标系中的最左下角的点,把根结点放到这个点中,然后对剩下的点进行极角排序,按逆时顺序一个个塞进来,类似地递归处理。

这样就满足了题意。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i) typedef long long LL; const int N = 1510; int X, Y; struct node{
int x, y;
int id;
friend bool operator < (const node &a, const node &b){
return (LL)(a.y - Y) * (b.x - X) < (LL)(b.y - Y) * (a.x - X);
}
} p[N]; vector <int> v[N];
int sz[N], ans[N];
int n; void dfs(int x, int fa){
sz[x] = 1;
for (auto u : v[x]){
if (u == fa) continue;
dfs(u, x);
sz[x] += sz[u];
}
} void calc(int x, int fa, int l, int r){
int t = l;
rep(i, l + 1, r){
if (p[i].y < p[t].y || (p[t].y == p[i].y && p[i].x < p[t].x))
t = i;
} if (t != l) swap(p[l], p[t]);
ans[p[l].id] = x;
X = p[l].x, Y = p[l].y;
sort(p + l + 1, p + r + 1);
int pos = l + 1; for (auto u : v[x]){
if (u == fa) continue;
calc(u, x, pos, pos + sz[u] - 1);
pos += sz[u];
}
} int main(){ scanf("%d", &n);
rep(i, 1, n - 1){
int x, y;
scanf("%d%d", &x, &y);
v[x].push_back(y);
v[y].push_back(x);
} rep(i, 1, n){
int x, y;
scanf("%d%d", &x, &y);
p[i] = {x, y, i};
} dfs(1, 0);
calc(1, 0, 1, n);
rep(i, 1, n) printf("%d\n", ans[i]); return 0;
}
05-16 21:30