Problem Statement
NOTE: There are images in the examples section of this problem statement that help describe the problem. Please view the problem statement in the HTML window to view them.
Given a picture composed entirely of horizontal and vertical line segments, calculate the minimum number of times you must lift your pen to trace every line segment in the picture exactly n times.
Each line segment will be of the form "<x1> <y1> <x2> <y2>" (quotes for clarity), representing a segment from (x1,y1) to (x2,y2). Segments may cross each other. Segments may also overlap, in which case you should count the overlapping region as appearing in the drawing only once. For example, say the drawing were composed of two lines: one from (6,4) to (9,4), and one from (8,4) to (14,4). Even though they overlap from (8,4) to (9,4), you should treat the drawing as if it were a single line from (6,4) to (14,4). You would not need to lift your pen at all to trace this drawing.
Definition
Class:PenLift
Method:numTimes
Parameters:String[], int
Returns:int
Method signature:int numTimes(String[] segments, int n)
(be sure your method is public)
Limits
Time limit (s):2.000
Memory limit (MB):64
Notes
-The pen starts on the paper at a location of your choice. This initial placement does not count toward the number of times that the pen needs to be lifted.
Constraints
-segments will contain between 1 and 50 elements, inclusive.
-Each element of segments will contain between 7 and 50 characters, inclusive.
-Each element of segments will be in the format "<X1><Y1><X2>_<Y2>" (quotes for clarity). The underscore character represents exactly one space. The string will have no leading or trailing spaces.
-<X1>, <Y1>, <X2>, and <Y2> will each be between -1000000 and 1000000, inclusive, with no unnecessary leading zeroes.
-Each element of segments will represent a horizontal or vertical line segment. No line segment will reduce to a point.
-n will be between 1 and 1000000, inclusive.
Examples
(0)
{"-10 0 10 0","0 -10 0 10"}
1
Returns: 1
This picture looks like a plus sign centered at the origin. One way to trace this image is to start your pen at (-10,0), move right to (10,0), lift your pen and place it at (0,-10), and then move up to (0,10). There is no way to trace the picture without lifting your pen at all, so the method returns 1.
(1)
{"-10 0 0 0","0 0 10 0","0 -10 0 0","0 0 0 10"}
1
Returns: 1
The picture is the same as the previous example, except that it has been described with four line segments instead of two. Therefore, the method still returns 1.
(2)
{"-10 0 0 0","0 0 10 0","0 -10 0 0","0 0 0 10"}
4
Returns: 0
You are now required to trace each segment exactly 4 times. You can do so without lifting your pen at all. Start at (0,0). Move your pen left to (-10,0), then back right to (0,0), then left again to (-10,0), then right again to (0,0). You have now traced the first line segment 4 times. Repeat this process for the other three segments as well. Since no pen lifts were required, the method returns 0.
(3)
{"0 0 1 0", "2 0 4 0", "5 0 8 0", "9 0 13 0",
"0 1 1 1", "2 1 4 1", "5 1 8 1", "9 1 13 1",
"0 0 0 1", "1 0 1 1", "2 0 2 1", "3 0 3 1",
"4 0 4 1", "5 0 5 1", "6 0 6 1", "7 0 7 1",
"8 0 8 1", "9 0 9 1", "10 0 10 1", "11 0 11 1",
"12 0 12 1", "13 0 13 1"}
1
Returns: 6
The picture looks like this:
To trace the picture using the minimum number of pen lifts, refer to the following diagram:
Start by placing your pen at the yellow dot. Trace the yellow square. Now lift your pen and place it on the red dot. Move downward, tracing the vertical line segment, and then around the perimeter of the red rectangle. Lift your pen again and place it on the green dot. Trace the green lines using the same method as you did for the red lines. Lift your pen a third time, placing it on the magenta dot. Trace the magenta lines in a similar fashion. You will need to lift your pen three more times to trace each of the leftover white segments, for a grand total of 6 pen lifts.
(4)
{"-2 6 -2 1", "2 6 2 1", "6 -2 1 -2", "6 2 1 2",
"-2 5 -2 -1", "2 5 2 -1", "5 -2 -1 -2", "5 2 -1 2",
"-2 1 -2 -5", "2 1 2 -5", "1 -2 -5 -2", "1 2 -5 2",
"-2 -1 -2 -6","2 -1 2 -6","-1 -2 -6 -2","-1 2 -6 2"}
5
Returns: 3
This is an example of overlap. Once all the segments are drawn, the picture looks like this:
You would need to lift your pen 3 times to trace every segment in this drawing exactly 5 times.
(5)
{"-252927 -1000000 -252927 549481","628981 580961 -971965 580961",
"159038 -171934 159038 -420875","159038 923907 159038 418077",
"1000000 1000000 -909294 1000000","1000000 -420875 1000000 66849",
"1000000 -171934 628981 -171934","411096 66849 411096 -420875",
"-1000000 -420875 -396104 -420875","1000000 1000000 159038 1000000",
"411096 66849 411096 521448","-971965 580961 -909294 580961",
"159038 66849 159038 -1000000","-971965 1000000 725240 1000000",
"-396104 -420875 -396104 -171934","-909294 521448 628981 521448",
"-909294 1000000 -909294 -1000000","628981 1000000 -909294 1000000",
"628981 418077 -396104 418077","-971965 -420875 159038 -420875",
"1000000 -1000000 -396104 -1000000","-971965 66849 159038 66849",
"-909294 418077 1000000 418077","-909294 418077 411096 418077",
"725240 521448 725240 418077","-252927 -1000000 -1000000 -1000000",
"411096 549481 -1000000 549481","628981 -171934 628981 923907",
"-1000000 66849 -1000000 521448","-396104 66849 -396104 1000000",
"628981 -1000000 628981 521448","-971965 521448 -396104 521448",
"-1000000 418077 1000000 418077","-1000000 521448 -252927 521448",
"725240 -420875 725240 -1000000","-1000000 549481 -1000000 -420875",
"159038 521448 -396104 521448","-1000000 521448 -252927 521448",
"628981 580961 628981 549481","628981 -1000000 628981 521448",
"1000000 66849 1000000 -171934","-396104 66849 159038 66849",
"1000000 66849 -396104 66849","628981 1000000 628981 521448",
"-252927 923907 -252927 580961","1000000 549481 -971965 549481",
"-909294 66849 628981 66849","-252927 418077 628981 418077",
"159038 -171934 -909294 -171934","-252927 549481 159038 549481"}
824759
Returns: 19
分析
本题实际上就是一笔画问题，可参考 http://www.jianshu.com/p/07771307c0bf
代码实现
import java.util.ArrayList;
import java.util.Collections;
public class PenLift {
// 测试方法
public static void main(String[] args) {
int num;
// // example 0
// String[] segs = {"-10 0 10 0", "0 -10 0 10"};
// num = 1;
// // example 1,2
// String[] segs = {"-10 0 0 0", "0 0 10 0", "0 -10 0 0", "0 0 0 10"};
// num = 1;
// num = 4;
// // example 3
// String[] segs = {"0 0 1 0", "2 0 4 0", "5 0 8 0", "9 0 13 0",
// "0 1 1 1", "2 1 4 1", "5 1 8 1", "9 1 13 1",
// "0 0 0 1", "1 0 1 1", "2 0 2 1", "3 0 3 1",
// "4 0 4 1", "5 0 5 1", "6 0 6 1", "7 0 7 1",
// "8 0 8 1", "9 0 9 1", "10 0 10 1", "11 0 11 1",
// "12 0 12 1", "13 0 13 1"};
// num = 1;
// example 4
String[] segs = {"-2 6 -2 1", "2 6 2 1", "6 -2 1 -2", "6 2 1 2",
"-2 5 -2 -1", "2 5 2 -1", "5 -2 -1 -2", "5 2 -1 2",
"-2 1 -2 -5", "2 1 2 -5", "1 -2 -5 -2", "1 2 -5 2",
"-2 -1 -2 -6","2 -1 2 -6","-1 -2 -6 -2","-1 2 -6 2"};
num = 5;
System.out.println(numTimes(segs, num));
}
static ArrayList<Point> points;
public static int numTimes(String[] segments, int n) {
ArrayList<Segment> segs = new ArrayList<Segment>();
for (int i = 0; i < segments.length; i++) {
String str[] = segments[i].split(" ");
int x1 = Integer.parseInt(str[0]);
int y1 = Integer.parseInt(str[1]);
int x2 = Integer.parseInt(str[2]);
int y2 = Integer.parseInt(str[3]);
segs.add(new Segment(x1, y1, x2, y2));
}
for (int i = 0; i < segs.size(); i++) {
Segment seg1 = segs.get(i);
for (int j = i + 1; j<segs.size(); j++) {
Segment seg2 = segs.get(j);
if (Segment.isOverlap(seg1, seg2) == true) {
// 将两条重合的线段，合并成一条
segs.set(i, Segment.combine(seg1, seg2));
segs.remove(j);
i--;
break;
}
}
}
points = new ArrayList<Point>();
// 把所有线段的端点、交点添加到points中
for (int i = 0; i < segs.size(); i++) {
Segment seg = segs.get(i);
Point p1 = new Point(seg.x1, seg.y1);
if (points.indexOf(p1) == -1) {
points.add(p1);
}
Point p2 = new Point(seg.x2, seg.y2);
if (points.indexOf(p2) == -1) {
points.add(p2);
}
for (int j = 0; j < i; j++) {
Segment seg1 = segs.get(j);
Point cross = Segment.intersect(seg, seg1);
if (cross != null) {
if (points.indexOf(cross) == -1) {
points.add(cross);
}
}
}
}
// 把points中的所有结点，根据连通关系放到不同的line中
for (int i = 0; i < segs.size(); i++) {
Segment seg = segs.get(i);
ArrayList<Point> line = new ArrayList<Point>();
for (int j = 0; j < points.size(); j++) {
// 同一条线上有哪些点
Point point = points.get(j);
if (point.x >= seg.x1 && point.x <= seg.x2 && point.y >= seg.y1 && point.y <= seg.y2) {
line.add(point);
}
}
Collections.sort(line);
for (int j = 0; j < line.size() - 1; j++) {
line.get(j).adjacents.add(points.indexOf(line.get(j + 1)));
ArrayList<Integer> ids = line.get(j).adjacents;
line.get(j + 1).adjacents.add(points.indexOf(line.get(j)));
ArrayList<Integer> ids2 = line.get(j+1).adjacents;
}
}
int count = 0;
// 求奇点个数
for (int i = 0; i < points.size(); i++) {
if (points.get(i).visited == false) {
points.get(i).visited = true;
int odd = DFS(i);
// 若n为偶数，points.get(i)必然不是奇点，可以一笔画
if (n % 2 == 0 || odd == 0) {
count++;
} else {
count += odd / 2;
}
}
}
return count - 1;
}
// 深度优先搜索
private static int DFS(int index) {
Point point = points.get(index);
int odd;
if (point.adjacents.size() % 2 == 0) {
odd = 0;
} else {
odd = 1;
}
// 累加points.get(index)所有相邻点的奇点数量
for (int i = 0; i < point.adjacents.size(); i++) {
int next = point.adjacents.get(i);
if (points.get(next).visited == false) {
points.get(next).visited = true;
odd += DFS(next);
}
}
return odd;
}
// 内部类：线段
static class Segment {
int x1;
int y1;
int x2;
int y2;
boolean horizontal;
public Segment(int X1, int Y1, int X2, int Y2) {
this.x1 = Math.min(X1, X2);
this.y1 = Math.min(Y1, Y2);
this.x2 = Math.max(X1, X2);
this.y2 = Math.max(Y1, Y2);
if (this.y1 == this.y2) {
this.horizontal = true;
} else {
this.horizontal = false;
}
}
// 判断两条线段是否重叠
static boolean isOverlap(Segment a, Segment b) {
if (a.horizontal == true && b.horizontal == true && a.y1 == b.y1) {
// 1. 两条都是水平线，且纵坐标相同
if ((a.x1 >= b.x1 && a.x1 <= b.x2) || (a.x2 >= b.x1 && a.x2 <= b.x2) || (b.x1 >= a.x1 && b.x1 <= a.x2) || (b.x2 >= a.x1 && b.x2 <= a.x2)) {
return true;
}
} else if (a.horizontal == false && b.horizontal == false && a.x1 == b.x2) {
// 2. 两条都是垂直线，且横坐标相同
if ((a.y1 >= b.y1 && a.y1 <= b.y2) || (a.y2 >= b.y1 && a.y2 <= b.y2) || (b.y1 >= a.y1 && b.y1 <= a.y2) || (b.y2 >= a.y1 && b.y2 <= a.y2)) {
return true;
}
}
return false;
}
// 联合两条线段，变成一条长的线段
static Segment combine(Segment a, Segment b) {
if (a.horizontal == true && b.horizontal == true) {
int x1 = Math.min(a.x1, b.x1);
int x2 = Math.max(a.x2, b.x2);
int y = a.y1;
return new Segment(x1, y, x2, y);
} else {
int x = a.x1;
int y1 = Math.min(a.y1, b.y1);
int y2 = Math.max(a.y2, b.y2);
return new Segment(x, y1, x, y2);
}
}
// 求两条相交线的交点
static Point intersect(Segment a, Segment b) {
// 若两条线段都为水平或垂直，则不可能相交
if (a.horizontal == b.horizontal) {
return null;
}
// 若a为垂直线b为水平线，交换两条线，使得a为水平线b为垂直线
if (a.horizontal == false) {
Segment temp = a;
a = b;
b = temp;
}
// a为水平线b为垂直线，只有相交的情况下，才有交点
if (a.y1 >= b.y1 && a.y1 <= b.y2 && b.x1 >= a.x1 && b.x1 <= a.x2) {
return new Point(b.x1,a.y1);
}
return null;
}
}
// 内部类：点
static class Point implements Comparable<Point> {
int x;
int y;
ArrayList<Integer> adjacents;
boolean visited;
public Point(int X, int Y) {
this.x = X;
this.y = Y;
this.adjacents = new ArrayList<Integer>();
this.visited = false;
}
public boolean equals(Object obj) {
Point another = (Point) obj;
if (this.x == another.x && this.y == another.y) {
return true;
} else {
return false;
}
}
public int compareTo(Point another) {
// 先按横坐标升序，若横坐标相同，按纵坐标升序
if (this.x != another.x) {
return this.x - another.x;
} else {
return this.y - another.y;
}
}
}
}
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