挑战程序竞赛系列（84）：3.6极限情况（1）
传送门：POJ 1981: Circle and Points
题意：
平面上有N个点，给定半径为1的圆，最多能在圆内点的个数。
这种题目很大特色在于，如果枚举圆的圆心，那么在偌大的空间中，有无数个圆，显然是不现实的。所以得考虑极限情况，也就是找出一种特殊的状态，更新它们的集合，能够获得答案。
此题的极限情况为：当两个点在一个圆上时，最大值一定在于这种情况。因此，我们可以枚举两个点构成的圆，并且计算该圆内有多少个点，更新最大值即可。
证明：
假设存在一个圆，圆内包含了最大个数的点，稍微移动圆的圆心不会发生任何变化，直到移动至出现两个点在圆上时，最大值依旧没有发生变化，但一旦再移动时，最大个数将减小。所以最大值一定存在于某两个点i和j构成的圆，且i和j在圆上or圆内。
参考至：
http://www.hankcs.com/program/algorithm/poj-1981-circle-and-points.html
所以我们只需要计算出两个点构成的极角，当然都得以i为坐标原点。
- atan2的角度为j和i的纵坐标与横坐标之比，取反tan。
- acos的角度为d/2，图中很清楚了。
- 所以两个极限角即为atan2 - acos和atan2 + acos。
最后，对极角排个序，遇到起始点加1，遇到终点减1，不断更新重叠区域个数的最大值即可。
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/P1981.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
static final int MAX_N = 302;
class P {
double x;
double y;
P(double x, double y){
this.x = x;
this.y = y;
}
}
class PolarAngle implements Comparable<PolarAngle>{
double angle;
boolean flag;
PolarAngle(double angle, boolean flag){
this.angle = angle;
this.flag = flag;
}
@Override
public int compareTo(PolarAngle that) {
return Double.compare(this.angle, that.angle);
}
}
P[] points = new P[MAX_N];
PolarAngle[] pa = new PolarAngle[2 * MAX_N];
int N;
int ans;
double dist(P a, P b) {
double dx = a.x - b.x;
double dy = a.y - b.y;
return Math.sqrt(dx * dx + dy * dy);
}
void solve() {
ans = 1;
for (int i = 0; i < N; ++i) {
int m = 0;
for (int j = 0; j < N; ++j) {
double d = 0;
if (i != j && (d = dist(points[i], points[j])) <= 2.0) {
double theta = Math.acos(d / 2);
double phi = Math.atan2(points[j].y - points[i].y, points[j].x - points[i].x);
pa[m++] = new PolarAngle(phi - theta, true); // start
pa[m++] = new PolarAngle(phi + theta, false); // end;
}
}
Arrays.sort(pa, 0, m);
int sum = 1;
for (int j = 0; j < m; ++j) {
if (pa[j].flag) {
sum ++;
}
else {
sum --;
}
ans = Math.max(ans, sum);
}
}
}
void read() {
while (true) {
N = ni();
if (N == 0) break;
for (int i = 0; i < N; ++i) {
points[i] = new P(nd(), nd());
}
ans = 0;
solve();
out.println(ans);
}
}
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);
}
}
}
