挑战程序竞赛系列（83）：3.6计算几何基础
传送门：POJ 1127: Jack Straws
之前计算几何这一块还未学习，今天开始把它们补上。
题意：
桌上放着n根木棍，木棍i的两端的坐标分别是(pix,piy)和(qix,qiy)(p_{ix}, p_{iy})和(q_{ix}, q_{iy})。给定m对木棍(ai,bi)(a_i, b_i)，请判断每对木棍是否相连。当两根木棍之间有公共点时，就认为它们是相连的。通过相连的木棍间接的连在一起的两根木棍也认为是相连的。
思路： 因为边和边是否相连就看交点是否在线段内，可以把每条线段想象成图中的顶点，只要有交点，就认为可达，最后判断任意两条线段是否相交，只需要判断它们是否可达。
所以问题就转换成了线段与线段相交的判断。分为两种情况：
- 边平行，需要判断任何一条线段的两个顶点是否在另一条线段上。
- 非平行边，求出两条线段的交点，判断交点是否分别在这两条线段内。
求外积，其实是求三点是否能够构成三角形，如果三角形的面积为0，说明三点共线。内积判断点是否在线段内，是因为如果向量夹角超过90度，内积为负。而点在线段内，向量的夹角一定为180度。
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.Map;
import java.util.StringTokenizer;
public class Main{
String INPUT = "./data/judge/201709/P1127.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
class P {
static final double EPS = 1e-10;
double x;
double y;
P(double x, double y){
this.x = x;
this.y = y;
}
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));
}
P mul(P a) {
return new P(x * a.x, y * a.y);
}
double dot(P a) {
return add(x * a.x, y * a.y);
}
double det(P a) {
return add(x * a.y, -y * a.x);
}
@Override
public String toString() {
return "(" x "," y ")";
}
public double add(double a, double b) {
if (Math.abs(a b) < EPS * (Math.abs(a) Math.abs(b))) return 0;
return a b;
}
}
boolean onSeg(P p1, P p2, P q) {
return p1.sub(q).det(p2.sub(q)) == 0 && p1.sub(q).dot(p2.sub(q)) <= 0;
}
P intersection(P p1, P p2, P q1, P q2) {
double fz = q2.sub(q1).det(q1.sub(p1));
double fm = q2.sub(q1).det(p2.sub(p1));
P q = p2.sub(p1);
q = new P(fz / fm * q.x, fz / fm * q.y);
return p1.add(q);
}
void read() {
while (true) {
int N = ni();
if (N == 0) break;
P[] p = new P[N];
P[] q = new P[N];
for (int i = 0; i < N; i) {
p[i] = new P(nd(), nd());
q[i] = new P(nd(), nd());
}
boolean[][] map = new boolean[N][N];
for (int i = 0; i < N; i) {
map[i][i] = true;
for (int j = i 1; j < N; j) {
if (p[i].sub(q[i]).det(p[j].sub(q[j])) == 0) {
map[i][j] = map[j][i] = onSeg(p[i], q[i], p[j])
|| onSeg(p[i], q[i], q[j])
|| onSeg(p[j], q[j], p[i])
|| onSeg(p[j], q[j], q[i]);
}
else {
P r = intersection(p[i], q[i], p[j], q[j]);
map[i][j] = map[j][i] = onSeg(p[i], q[i], r) && onSeg(p[j], q[j], r);
}
}
}
for (int k = 0; k < N; k) {
for (int i = 0; i < N; i) {
for (int j = 0; j < N; j) {
map[i][j] |= map[i][k] && map[k][j];
}
}
}
while (true) {
int a = ni();
int b = ni();
if (a b == 0) break;
a --;
b --;
out.println(map[a][b] ? "CONNECTED" : "NOT CONNECTED");
}
}
}
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);
}
}
}
