[codeforces934E]A Colourful Prospect

试题描述

Firecrackers scare Nian the monster, but they're wayyyyy too noisy! Maybe fireworks make a nice complement.

Little Tommy is watching a firework show. As circular shapes spread across the sky, a splendid view unfolds on the night of Lunar New Year's eve.

A wonder strikes Tommy. How many regions are formed by the circles on the sky? We consider the sky as a flat plane. A region is a connected part of the plane with positive area, whose bound consists of parts of bounds of the circles and is a curve or several curves without self-intersections, and that does not contain any curve other than its boundaries. Note that exactly one of the regions extends infinitely.

求 \(n\) 个圆将平面划分成的区域个数。

输入

The first line of input contains one integer \(n\) \((1 \le n \le 3)\), denoting the number of circles.

The following \(n\) lines each contains three space-separated integers \(x\), \(y\) and \(r\) \(( - 10 \le x, y \le 10, 1 \le r \le 10)\), describing a circle whose center is \((x, y)\) and the radius is \(r\). No two circles have the same \(x\), \(y\) and \(r\) at the same time.

输出

Print a single integer — the number of regions on the plane.

输入示例1

输出示例1

输入示例2

输出示例2

输入示例3

输出示例3

数据规模及约定

见“输入”

题解

这题要用到平面上的欧拉公式：\(V - E + R = C + 1\)，其中 \(V\)(vertex) 表示交点数目，\(E\)(edge) 表示边数，\(R\)(region) 表示区域数，\(C\)(connection) 用来分割的线所构成的连通块个数。（英文是我自己 YY 的）

然后这题就好做了，\(V\) 非常好求，两两圆求一下交点去重即可；\(E\) 就是每个圆上的交点个数之和（注意如果一个圆不与任何圆相交，那么它不能算作一条边）；\(C\) 也可以用并查集啥的统计一下；于是我们就算出 \(R\) 了。

分类讨论的话可能这辈子做不完这道题了。而且 \(n\) 可以出到 \(1000\) 级别

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <algorithm>

#include <cmath>

#include <vector>

using namespace std;

#define rep(i, s, t) for(int i = (s), mi = (t); i <= mi; i++)

#define dwn(i, s, t) for(int i = (s), mi = (t); i >= mi; i--)

int read() {

	int x = 0, f = 1; char c = getchar();

	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }

	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }

	return x * f;

}

const double eps = 1e-9;

struct Vector {

	double x, y;

	Vector() {}

	Vector(double _, double __): x(_), y(__) {}

	Vector operator - (const Vector& t) const { return Vector(x - t.x, y - t.y); }

	double len() { return sqrt(x * x + y * y); }

	bool operator < (const Vector& t) const { return abs(x - t.x) > eps ? x < t.x : y < t.y; }

	bool operator == (const Vector& t) const { return abs(x - t.x) <= eps && abs(y - t.y) <= eps; }

} ps[100], crs[5][100];

int cp, cc[5];

struct Circle {

	Vector p; double r;

	Circle() {}

	Circle(Vector _, double __): p(_), r(__) {}

	Vector point(double a) { return Vector(cos(a) * r + p.x, sin(a) * r + p.y); }

} cs[5];

double angle(Vector x) { return atan2(x.y, x.x); }

vector <Vector> getcross(Circle c1, Circle c2) {

	vector <Vector> p; p.resize(0);

	Vector t = c2.p - c1.p;

	double d = t.len(), a = angle(t);

	if(abs(d) <= eps) return p;

	if(d < abs(c2.r - c1.r) || d > c2.r + c1.r) return p;

	double da = acos((c1.r * c1.r + d * d - c2.r * c2.r) / (2 * c1.r * d));

	Vector p1 = c1.point(a + da), p2 = c1.point(a - da);

	p.push_back(p1);

	if(p1 == p2) return p;

	p.push_back(p2);

	return p;

}

int fa[5];

int findset(int x) { return x == fa[x] ? x : fa[x] = findset(fa[x]); }

int main() {

	int n = read();

	rep(i, 1, n) {

		int x = read(), y = read(), r = read();

		cs[i] = Circle(Vector(x, y), r);

		fa[i] = i;

	}

	int V, E = 0, C = n;

	rep(i, 1, n) {

		rep(j, 1, n) if(i != j) {

			vector <Vector> tmp = getcross(cs[i], cs[j]);

			for(auto k : tmp) crs[i][++cc[i]] = k;

			if(tmp.size() > 0) {

				int u = findset(i), v = findset(j);

				if(u != v) fa[v] = u, C--;

			}

		}

		sort(crs[i] + 1, crs[i] + cc[i] + 1);

		cc[i] = unique(crs[i] + 1, crs[i] + cc[i] + 1) - crs[i] - 1;

		E += cc[i];

		/*printf("cs[%d]:\n", i);

		rep(j, 1, cc[i]) printf("(%.5lf, %.5lf)\n", crs[i][j].x, crs[i][j].y); // */

		rep(j, 1, cc[i]) ps[++cp] = crs[i][j];

	}

	sort(ps + 1, ps + cp + 1);

	cp = unique(ps + 1, ps + cp + 1) - ps - 1;

	V = cp;

	// printf("E: %d, V: %d, C: %d\n", E, V, C);

	printf("%d\n", E - V + C + 1); // */

	return 0;

}