East Central North America 2004 I Conduit!

【题目描述】

题目意思不难理解，把n条线段画在纸上，合并可以合并的线段，比如两条线段有至少两个公共点就可以合并成一条线段，问最后剩下几条线段

【输入输出】

多case

n (n <= 10000)

x1, y1, x12, y12

....

xn, yn, xn2, yn2;

n = 0结束

每个用例输出最后的线段数。

【题目分析】

想法就是对线段按斜率排序，斜率相同的按x轴或y轴的截距来排序，截距也相同的按x较小的x坐标或者y较小的y坐标来排序，这样就可以保证overlap的线段都是相邻的，然后线性扫描一遍。

注意水平线和垂线的特殊处理

【code】

 /** * */ import java.util.Arrays; import java.util.Scanner; /** * @author 张国文 * */ public class Main { /** * @param args */ public static void main(String[] args) { new Main().start(); } class Point { Point(double x, double y) { this.x = x; this.y = y; } double x, y; } class Segment implements Comparable<Segment> { Segment(Point a, Point b) { this.a = a; this.b = b; if (a.x == b.x) { if (a.y > b.y) { swap(); } } else if (a.x > b.x) { swap(); } } private void swap() { Point c = a; a = b; b = c; } Point a, b; @Override public int compareTo(Segment that) { double dy1 = this.b.y - this.a.y; double dx1 = this.b.x - this.a.x; double dy2 = that.b.y - that.a.y; double dx2 = that.b.x - that.a.x; if (same(dy1 * dx2, dy2 * dx1)) { if (same(this.a.x, this.b.x)) { if (same(this.a.x, that.a.x)) { return this.a.y < that.a.y ? -1 : 1; } return this.a.x < that.a.x ? -1 : 1; } if (same(this.a.y, this.b.y)) { if (same(this.a.y, that.a.y)) { return this.a.x < that.a.x ? -1 : 1; } return this.a.y < that.a.y ? -1 : 1; } if (same((this.a.x * this.b.y - this.a.y * this.b.x) * (that.a.x - that.b.x), (that.a.x * that.b.y - that.a.y * that.b.x) * (this.a.x - this.b.x))) { return this.a.x < that.a.x ? -1 : 1; } return (this.a.x * this.b.y - this.a.y * this.b.x) * (that.a.x - that.b.x) < (that.a.x * that.b.y - that.a.y * that.b.x) * (this.a.x - this.b.x) ? -1 : 1; } return dy1 * dx2 < dy2 * dx1 ? -1 : 1; } } private boolean same(double a, double b) { return Math.abs(a - b) < 1E-7; } private void start() { Scanner in = new Scanner(System.in); int n; Segment seg[]; while ((n = in.nextInt()) != 0) { seg = new Segment[n]; for (int i = 0; i < n; i++) { seg[i] = new Segment( new Point(in.nextDouble(), in.nextDouble()), new Point( in.nextDouble(), in.nextDouble())); } Arrays.sort(seg); int cnt = 0; for (int i = 0; i < n;) { if (same(seg[i].a.x, seg[i].b.x)) { double right = seg[i].b.y; int j = i; while (j < n && same(seg[j].a.x, seg[i].a.x) && seg[j].a.y <= right) { if (seg[j].b.y > right) { right = seg[j].b.y; } j++; } cnt++; i = j; continue; } double right = seg[i].b.x; int j = i; while (j < n && same((seg[j].b.y - seg[j].a.y) * (seg[i].b.x - seg[i].a.x), (seg[i].b.y - seg[i].a.y) * (seg[j].b.x - seg[j].a.x)) && same((seg[i].a.x * seg[i].b.y - seg[i].a.y * seg[i].b.x) * (seg[j].a.x - seg[j].b.x), (seg[j].a.x * seg[j].b.y - seg[j].a.y * seg[j].b.x) * (seg[i].a.x - seg[i].b.x)) && seg[j].a.x <= right) { if (seg[j].b.x > right) { right = seg[j].b.x; } j++; } cnt++; i = j; } System.out.println(cnt); } } }