挑战程序竞赛系列（90）：3.6凸包（1）

挑战程序竞赛系列（90）：3.6凸包（1）

传送门：POJ 2187: Beauty Contest

题意：

平面上有N个牧场。i号牧场的位置在格点(xi,yi)，所有牧场的位置互不相同。请计算距离最远的两个牧场的距离，输出最远距离的平方。

假设有四个点，其中一个点在三个点的内部，可以知道该点是冗余的，所以我们只需要维护所有点的凸包，求出凸包上点集中的两两最大即可。

凸包的详细介绍可以参考博文：

http://blog.csdn.net/u014688145/article/details/72200018#t3

暴力枚举点对

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.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class Main{

String INPUT = "./data/judge/201709/P2187.txt";

public static void main(String[] args) throws IOException {
new Main().run();
}

static final int MAX_N = 50000;

class Point implements Comparable<Point>{
int x;
int y;

Point(int x, int y){
this.x = x;
this.y = y;
}

Point sub(Point a) {
return new Point(x - a.x, y - a.y);
}

int det(Point a) {
return x * a.y - y * a.x;
}

@Override
public int compareTo(Point o) {
return this.x != o.x ? this.x - o.x : this.y - o.y;
}
}

int N;
Point[] ps;

List<Point> convexHull(){
Arrays.sort(ps);

List<Point> ans = new ArrayList<Point>();

int[] stack = new int[2 * N];
int tot = 0;
for (int i = 0; i < N; ++i) {
while (tot > 1 && ps[stack[tot - 1]].sub(ps[stack[tot - 2]]).det(ps[i].sub(ps[stack[tot - 1]])) <= 0) tot--;
stack[tot++] = i;
}

for (int i = N - 2, t = tot; i >= 0; --i) {
while (tot > t && ps[stack[tot - 1]].sub(ps[stack[tot - 2]]).det(ps[i].sub(ps[stack[tot - 1]])) <= 0) tot--;
stack[tot++] = i;
}
tot --;

for (int i = 0; i < tot; ++i) {
ans.add(ps[stack[i]]);
}

return ans;
}

int dist(Point a, Point b) {
int dx = a.x - b.x;
int dy = a.y - b.y;
return dx * dx + dy * dy;
}

void solve() {

List<Point> qs = convexHull();
int max = 0;
for (int i = 0; i < qs.size(); ++i) {
for (int j = i + 1; j < qs.size(); ++j) {
max = Math.max(max, dist(qs.get(i), qs.get(j)));
}
}
out.println(max);
}

void read() {
N = ni();
ps = new Point
;
for (int i = 0; i < N; ++i) {
ps[i] = new Point(ni(), ni());
}
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);
}
}
}

旋转卡壳法

参考博文：

http://www.cnblogs.com/DreamUp/archive/2010/09/16/1828131.html

参考《挑战》P263，借用书上的解释，和上篇博文的一张图：



实际上，最远点对一定在所有对踵点对上，所以只需要枚举出凸包上的对踵点对即能高效求解最远点对。

那么假设我们知道了对踵点对，qa和qb，如何求下一个对踵点对呢？考虑qa和qa的下一个顶点，qb和qb的下一顶点，可以画出两条向量，并且以qa为原点，做qb和qb的下一顶点的平行线，如下：



于是根据叉积可以得到两向量的相对位置关系（谁在最外侧），如果平移后的线在内侧，则说明对踵点对为qb和qa的next，反之为qb的next和qa。

有了上述关系，就可以直接根据对踵点对更新最远点对了，而我们知道在对踵点对内部的点是不可能为最远点对的，这是复杂度降低的真正原因。

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/201710/P2187.txt";

public static void main(String[] args) throws IOException {
new Main().run();
}

static final int MAX_N = 50000 + 16;
int N;

class P implements Comparable<P>{
int x;
int y;

P(int x, int y){
this.x = x;
this.y = y;
}

P sub(P a) {
return new P(x - a.x, y - a.y);
}

int det(P a) {
return x * a.y - y * a.x;
}

@Override
public int compareTo(P o) {
return x != o.x ? x - o.x : y - o.y;
}
}

P[] ps;

int dist(P a, P b) {
int dx = a.x - b.x;
int dy = a.y - b.y;
return dx * dx + dy * dy;
}

P[] convexHull() {
P[] qs = new P[2 * N];
Arrays.sort(ps);
int k = 0;
for (int i = 0; i < N; ++i) {
while (k > 1 && qs[k - 1].sub(qs[k - 2]).det(ps[i].sub(qs[k - 1])) <= 0) k--;
qs[k++] = ps[i];
}

for (int i = N - 2, t = k; i >= 0; --i) {
while (k > t && qs[k - 1].sub(qs[k - 2]).det(ps[i].sub(qs[k - 1])) <= 0) k--;
qs[k++] = ps[i];
}

k --;
P[] res = new P[k];
System.arraycopy(qs, 0, res, 0, k);
return res;
}

void solve() {
P[] qs = convexHull();
int n = qs.length;
if (n == 2) {
out.println(dist(qs[0], qs[1]));
return;
}

int i = 0;
int j = 0;
for (int k = 1; k < n; ++k) {
if (qs[k].x < qs[i].x) i = k;
if (qs[k].x > qs[j].x) j = k;
}

int si = i;
int sj = j;

int max = 0;
while (i != sj || j != si) {
max = Math.max(max, dist(qs[i], qs[j]));
if (qs[(i + 1) % n].sub(qs[i]).det(qs[(j + 1) % n].sub(qs[j])) < 0) {
i = (i + 1) % n;
}
else {
j = (j + 1) % n;
}
}
out.println(max);
}

void read() {
N = ni();
ps = new P
;
for (int i = 0; i < N; ++i) {
ps[i] = new P(ni(), ni());
}
solve();
}

FastScanner in;
PrintWriter out;

void run() throws IOException {
boolean oj;
try {
oj = ! System.getProperty("user.dir").equals("F:\\oxygen_workspace\\Algorithm");
} 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);
}
}
}