题意:
礼花:Confetti 是一些大小不一的彩色圆形纸片,人们在派对上、过节时便抛洒它们以示庆祝。落在地上的Confetti会堆叠起来,以至于一部分会被盖住而看不见。给定Confetti的尺寸和位置以及它们的叠放次序,你能计算出有多少Confetti是可以看见的吗?
思路:
此题的确需要丰富的想象力,一开始简单的以为圆心距离小于两圆半径之差即可,其实它只是其中一种覆盖情况。如下图:
实际上,还可以有:
所以按照上述思路肯定会出现漏判的情况,那么该怎么办呢?参考神牛的思路:
如果底层的某个圆上的所有圆弧能够被上层的圆覆盖,则说明该底层圆是不可见的。的确涵盖了几乎所有的情况,但还是有特例哟!比如:
这种情况就需要做特殊处理了,想象一下,如果把最底层的绿色圆变大一些,或者变小一些,必然有些边不能被覆盖到。所以我们需要求出每段圆弧,并在此基础上扩大圆的半径,进行特判。对应代码中,t = -1 和 t = 1的循环。(具体参看代码)
接着分析可见与不可见的圆,因为我们对圆进行了离散化处理,实际是分析每段圆弧是否能找到对应的上层圆将它覆盖,如果在某一段圆弧中,搜遍了所有上层圆,都没能将一条弧覆盖,那么此底层圆必然是可见的。
在搜索底层圆的上层圆时,从上往下盖住的第一个圆也是可见的。
所以我们只需找到第一个盖住底层圆的上层圆即可跳出,如果找不到这样的圆,程序自然找的是它自己,因为自己经过扩张后,总能将自己覆盖。
此处就把上述两种情况合并在一块了,的确高级。
证明:(反证法)
假设第一个盖住底层圆的圆a不可见,那么必然被其上层的圆{c,d,e…}所覆盖,那么必然可以将圆a的弧分成若干段,分别找到最上层的圆{c,d,e…}将其覆盖,而我们知道圆a与底层圆的弧是最小划分单元,矛盾,得证。
弧的离散化:
和上篇博文求弧的思路一致,参考链接:
https://cloud.tencent.com/developer/article/1010099
代码如下:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main{
String INPUT = "./data/judge/201709/P1418.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
static final double PI = Math.acos(-1);
static final double EPS = 5E-13;
static final int MAX_N = 102;
class P{
double x;
double y;
P(double x, double y){
this.x = x;
this.y = y;
}
}
P[] o = new P[MAX_N]; // 圆心坐标
double[] r = new double[MAX_N]; // 圆半径
boolean[] visible = new boolean[MAX_N]; // 对应圆是否可见
int N;
double[] angle = new double[2 * MAX_N];
int tot;
double distance(P a, P b) {
double dx = a.x - b.x;
double dy = a.y - b.y;
return Math.sqrt(dx * dx + dy * dy);
}
double norm(double ang) {
while (ang < 0) {
ang += 2 * PI;
}
while (ang > 2 * PI) {
ang -= 2 * PI;
}
return ang;
}
void solve() {
Arrays.fill(visible, false);
for (int i = 0; i < N; ++i) {
tot = 0;
angle[tot++] = 0;
angle[tot++] = 2 * PI;
for (int j = 0; j < N; ++j) {
if (i == j) continue;
double d = distance(o[i], o[j]);
if (r[i] + r[j] < d || d < r[i] - r[j] || d < r[j] - r[i]) continue; // 包含或者不相交
double phi = Math.atan2(o[j].y - o[i].y, o[j].x - o[i].x);
double the = Math.acos((r[i] * r[i] + d * d - r[j] * r[j]) / (2 * r[i] * d));
angle[tot++] = norm(phi - the);
angle[tot++] = norm(phi + the);
}
Arrays.sort(angle, 0, tot);
for (int j = 0; j < tot - 1; ++j) {
double mid = (angle[j] + angle[j + 1]) / 2;
double nx = 0;
double ny = 0;
for (int t = -1; t < 2; t += 2) {
nx = o[i].x + (r[i] + t * EPS) * Math.cos(mid);
ny = o[i].y + (r[i] + t * EPS) * Math.sin(mid);
P np = new P(nx, ny);
int k = 0;
for (k = N - 1; k >= 0; --k) {
if (distance(np, o[k]) < r[k]) {
break;
}
}
if (k != -1)
visible[k] = true;
}
}
}
int ans = 0;
for (int i = 0; i < N; ++i) {
if (visible[i]) ans ++;
}
out.println(ans);
}
void read() {
while (true) {
N = ni();
if (N == 0) break;
for (int i = 0; i < N; ++i) {
o[i] = new P(nd(), nd());
r[i] = nd();
}
solve();
}
}
FastScanner in;
PrintWriter out;
void run() throws IOException {
boolean oj;
try {
oj = ! System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
} catch (Exception e) {
oj = System.getProperty("ONLINE_JUDGE") != null;
}
InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
in = new FastScanner(is);
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
read();
out.flush();
if (!oj){
System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
}
}
public boolean more(){
return in.hasNext();
}
public int ni(){
return in.nextInt();
}
public long nl(){
return in.nextLong();
}
public double nd(){
return in.nextDouble();
}
public String ns(){
return in.nextString();
}
public char nc(){
return in.nextChar();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
boolean hasNext;
public FastScanner(InputStream is) throws IOException {
br = new BufferedReader(new InputStreamReader(is));
hasNext = true;
}
public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
hasNext = false;
return "##";
}
}
return st.nextToken();
}
String next = null;
public boolean hasNext(){
next = nextToken();
return hasNext;
}
public int nextInt() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Integer.parseInt(more);
}
public long nextLong() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Long.parseLong(more);
}
public double nextDouble() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Double.parseDouble(more);
}
public String nextString(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more;
}
public char nextChar(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more.charAt(0);
}
}
}