挑战程序竞赛系列（89）：3.6平面扫描（3）
挑战程序竞赛系列（89）：3.6平面扫描（3）
传送门：POJ 3292: Rectilinear polygon
题意参考hankcs： http://www.hankcs.com/program/algorithm/poj-3293-rectilinear-polygon.html
题意：
直角多边形：给定N个点，问是否能组成直角多边形（每个顶点都与另外两个顶点构成直角，每条边都平行于坐标轴），并求出周长？
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思路参考： http://www.cnblogs.com/ZefengYao/p/7470984.html
平面扫描，按照x轴扫描可以获得所有竖边的长度和，按y轴的同理，先讨论x轴的情况，将点按照x坐标大小排序后，同一列上的点按照y坐标从小到大排序，之后再观察多边形每一列的特性，可以发现每一列上点必定偶数个，相邻两个配对可以成为一条边，若出现奇数条边，肯定是构不成多边形的。按y轴扫描也是相同做法。其次还要判断是否有横竖边相交的情况以及是否有洞(图是否连通)即可。
讲的很清楚了，此题的trick在于如何实现水平线与垂直线的相交判断，记录哪些信息可以检测出相交问题呢？
首先按照x轴扫描，不会出现相交的线，所以只需要把每条线段信息记录在一种数据结构即可，方便y轴扫描时判断相交。显然，给定一条水平线的横坐标的两端，我们只需要比较该区间内是否有垂直线与之相交，采用TreeSet维护x轴垂直线的有序，接着只要在扫描y轴时，知道两个端点，就可以拿到该区间内的垂直线逐个判断即可。
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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.Comparator;
import java.util.StringTokenizer;
import java.util.TreeSet;
public class Main{
String INPUT = "./data/judge/201709/P3293.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
static final int MAX_N = 100000 + 16;
class Point {
int x;
int y;
Point[] nxt;
int tot;
Point(int x, int y){
this.x = x;
this.y = y;
this.tot = 0;
this.nxt = new Point[2];
}
void add(Point a) {
nxt[tot++] = a;
}
@Override
public boolean equals(Object obj) {
Point p = (Point) obj;
return p.x == x && p.y == y;
}
@Override
public String toString() {
return "(" + x + " " + y + ")";
}
}
class Line implements Comparable<Line>{
int x;
int y0;
int y1;
Point[] ps;
Line (int x, int y0, int y1){
this.x = x;
this.y0 = y0;
this.y1 = y1;
ps = new Point[2];
}
int distance() {
int dx = ps[0].x - ps[1].x;
int dy = ps[0].y - ps[1].y;
return Math.abs(dx) + Math.abs(dy);
}
@Override
public int compareTo(Line o) {
return this.x - o.x;
}
@Override
public boolean equals(Object obj) {
Line o = (Line) obj;
return o.x == x && o.y0 == y0 && o.y1 == y1;
}
@Override
public String toString() {
return ps[0].toString() + "->" + ps[1].toString();
}
}
int N;
Point[] ps;
boolean intersect(Point a, Point b, Point c, Point d) { // 左 a 右 b 下 c 上 d
return a.x < c.x && b.x > c.x && c.y < a.y && d.y > a.y;
}
long solve() {
TreeSet<Line> lines = new TreeSet<Line>();
long ans = 0;
Arrays.sort(ps, 0, N, new Comparator<Point>() {
@Override
public int compare(Point o1, Point o2) {
return o1.x != o2.x ? o1.x - o2.x : o1.y - o2.y;
}
});
for (int i = 0; i < N; ++i) { // 按x轴方向扫描
int j = i;
while (j < N && ps[j].x == ps[i].x) ++j;
if (((j - i) & 1) != 0) return -1;
else {
for (int k = i; k < j; k += 2) {
ps[k].add(ps[k + 1]);
ps[k + 1].add(ps[k]);
Line line = new Line(ps[k].x, ps[k].y, ps[k + 1].y);
line.ps[0] = ps[k];
line.ps[1] = ps[k + 1];
ans += line.distance();
lines.add(line);
}
}
i = j - 1;
}
Arrays.sort(ps, 0, N, new Comparator<Point>() {
@Override
public int compare(Point o1, Point o2) {
return o1.y != o2.y ? o1.y - o2.y : o1.x - o2.x;
}
});
for (int i = 0; i < N; ++i) { //按y轴方向扫描
int j = i;
while (j < N && ps[j].y == ps[i].y) ++j;
if (((j - i) & 1) != 0) return -1;
else {
for (int k = i; k < j; k +=2) {
ps[k].add(ps[k + 1]);
ps[k + 1].add(ps[k]);
Line l1 = new Line(ps[k].x, 0, 0);
Line l2 = new Line(ps[k + 1].x, 0, 0);
l1.ps[0] = ps[k];
l1.ps[1] = ps[k + 1];
for (Line line : lines.subSet(l1, l2)) {
if (intersect(l1.ps[0], l1.ps[1], line.ps[0], line.ps[1])) return -1;
}
ans += l1.distance();
}
}
i = j - 1;
}
Point crt = ps[0];
Point pre = null;
int cnt = 0;
do {
cnt ++;
Point tmp = crt;
if (pre == crt.nxt[0]) {
crt = crt.nxt[1];
}
else{
crt = crt.nxt[0];
}
pre = tmp;
}
while (!crt.equals(ps[0]));
return cnt == N ? ans : -1;
}
void read() {
int T = ni();
while (T --> 0) {
N = ni();
ps= new Point[N];
for (int i = 0; i < N; ++i) {
int x = ni();
int y = ni();
ps[i] = new Point(x, y);
}
out.println(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);
}
}
}
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原始发表：2017-09-29 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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