挑战程序竞赛系列（92）：3.6凸包（3）
传送门：POJ 1912: A highway and the seven dwarfs
题意：
高铁与七个小矮人：侏儒岛上有N栋房子，组成一个社区。现给定一条高铁，问该高铁是否分割了社区？
此题不错，好题。很明显，判断社区是否被分割时，只需判断外围的点集即可（凸包上的点），因为凸包所能围成的区域最大，那么社区更容易被分割。
求完凸包上的点后，朴素的做法，暴力枚举每个点对，判断是否与直线相交，肯定超时，不写了，也没啥技术含量。
实际上，凸包上的很多点也是冗余的，对直线相交判断没有任何贡献，如何说？
看图：
实际上，给定一条直线，我们还能求出凸包中关于这条直线的最大间距，真正的目的在于max d，那么什么时候能够max d呢？不难想象，对应的凸包上的极角分别旋转一点即可，平行的时候达到极值。对应图中，我们要找到A2点和A4点。
而凸包在枚举每个点对的极角时，是自然有序的，但需要做一些处理，因为在求解A4A1对应的极角时，为负值需要加个2PI，当然A1A2也是负值，却不需要加，因为凸包的起点是按照x轴排序的，注意下细节就好。
代码如下：
代码语言：javascript复制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/P1912.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
static final int MAX_N = 100000 16;
static final double EPS = 1E-10;
static final double PI = Math.acos(-1.0);
class P implements Comparable<P>{
double x;
double y;
P (double x, double y){
this.x = x;
this.y = y;
}
double add(double a, double b) {
if (Math.abs(a b) < EPS * (Math.abs(a) Math.abs(b))) return 0;
return a b;
}
P add(P a) {
return new P(add(x, a.x), add(y, a.y));
}
P sub(P a) {
return new P(add(x, -a.x), add(y, -a.y));
}
double dot(P a) {
return a.x * x a.y * y;
}
double det(P a) {
return x * a.y - y * a.x;
}
@Override
public int compareTo(P o) {
int cmp = Double.compare(x, o.x);
return cmp != 0 ? cmp : Double.compare(y, o.y);
}
}
int N;
P[] ps;
P[] convexHull() {
Arrays.sort(ps);
if (N <= 1) return ps;
P[] qs = new P[N * 2];
int k = 0;
for (int i = 0; i < N; qs[k ] = ps[i ]) {
while (k > 1 && qs[k - 1].sub(qs[k - 2]).det(ps[i].sub(qs[k - 1])) < EPS) k--;
}
for (int i = N - 2, t = k; i >= 0; qs[k ] = ps[i--]) {
while (k > t && qs[k - 1].sub(qs[k - 2]).det(ps[i].sub(qs[k - 1])) < EPS) k--;
}
P[] res = new P[k - 1];
System.arraycopy(qs, 0, res, 0, k - 1);
return res;
}
double angle(P a, P b) {
P t = b.sub(a);
double ang = Math.atan2(t.y, t.x);
if (ang < -PI / 2 EPS) ang = 2 * PI;
return ang;
}
int lowerBound(double[] ds, double v) {
int l = 0, r = ds.length;
while (l < r) {
int m = (r l) >> 1;
if (ds[m] < v) l = m 1;
else r = m;
}
return l;
}
//a, b是否在直线st的两侧
boolean check(P a, P b, P s, P t) {
return a.sub(s).det(t.sub(s)) * b.sub(s).det(t.sub(s)) <= 0;
}
boolean cross(P[] qs, double[] ds, P p1, P p2) {
int i = lowerBound(ds, angle(p1, p2));
int j = lowerBound(ds, angle(p2, p1));
i %= qs.length;
j %= qs.length;
return check(qs[i], qs[j], p1, p2);
}
void solve() {
P[] qs = convexHull();
int n = qs.length;
double[] ds = new double[n];
for (int i = 0; i < n; i) {
ds[i] = angle(qs[i], qs[(i 1) % n]);
}
while (more()) {
P p1 = new P(nd(), nd());
P p2 = new P(nd(), nd());
if (N <= 1 || !cross(qs, ds, p1, p2)) out.println("GOOD");
else out.println("BAD");
}
}
void read() {
N = ni();
ps = new P[N];
for (int i = 0; i < N; i) {
ps[i] = new P(nd(), 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);
}
}
}
