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TA贡献1886条经验 获得超2个赞
基于以下假设进行更新:
问题中给出的第一行坐标表示连接线集中第一条线的坐标
找到第一组连接线后,问题中给出的剩余坐标线中第一条未使用的线将被视为下一组连接线中第一条线的坐标,依此类推,直到问题中给出的坐标列表为止问题排气。
所需的逻辑在Main.java
. 我还创建Point.java
并Line.java
测试了逻辑。如果您遇到任何问题,请随时告诉我。
Point.java
public class Point {
int x,y;
public Point(int x, int y) {
super();
this.x = x;
this.y = y;
}
public int getX() {
return x;
}
public void setX(int x) {
this.x = x;
}
public int getY() {
return y;
}
public void setY(int y) {
this.y = y;
}
@Override
public String toString() {
return "Point [x=" + x + ", y=" + y + "]";
}
}
Line.java
public class Line {
int x1,y1,x2,y2;
Point start,end;
boolean used;
public Line(int x1, int y1, int x2, int y2) {
super();
this.x1 = x1;
this.y1 = y1;
this.x2 = x2;
this.y2 = y2;
}
public Line(Point start, Point end) {
super();
this.start = start;
this.end = end;
}
public int getX1() {
return x1;
}
public void setX1(int x1) {
this.x1 = x1;
}
public int getY1() {
return y1;
}
public void setY1(int y1) {
this.y1 = y1;
}
public int getX2() {
return x2;
}
public void setX2(int x2) {
this.x2 = x2;
}
public int getY2() {
return y2;
}
public void setY2(int y2) {
this.y2 = y2;
}
public Point getStart() {
return start;
}
public void setStart(Point start) {
this.start = start;
}
public Point getEnd() {
return end;
}
public void setEnd(Point end) {
this.end = end;
}
public boolean isUsed() {
return used;
}
public void setUsed(boolean used) {
this.used = used;
}
@Override
public String toString() {
return "Line [x1=" + x1 + ", y1=" + y1 + ", x2=" + x2 + ", y2=" + y2 + "]";
}
}
Main.java
import java.util.ArrayList;
import java.util.List;
class Main {
public static void main(String args[]) {
List<Line> givenLines = new ArrayList<Line>();
givenLines.add(new Line(3350, 1500, 3200, 1500));
givenLines.add(new Line(1450, 1750, 1450, 2200));
givenLines.add(new Line(1450, 2200, 2100, 2200));
givenLines.add(new Line(2400, 2200, 2550, 2200));
givenLines.add(new Line(2550, 2200, 2550, 2350));
givenLines.add(new Line(2550, 2350, 2850, 2350));
givenLines.add(new Line(2850, 2350, 2850, 2700));
givenLines.add(new Line(2850, 2700, 3350, 2700));
givenLines.add(new Line(3650, 2700, 3750, 2700));
givenLines.add(new Line(3750, 2700, 3750, 2600));
givenLines.add(new Line(3750, 2600, 5250, 2600));
givenLines.add(new Line(5250, 2600, 5250, 2350));
givenLines.add(new Line(5250, 2350, 5000, 2350));
givenLines.add(new Line(4700, 2350, 4350, 2350));
givenLines.add(new Line(4350, 2350, 4350, 1600));
givenLines.add(new Line(4350, 1600, 3650, 1600));
givenLines.add(new Line(3650, 1600, 3650, 1500));
givenLines.add(new Line(3200, 1500, 3200, 1750));
givenLines.add(new Line(3200, 1750, 1450, 1750));
int linesIndex, usedCounter=0;
List<List<Line>> listOfConnectedLines = new ArrayList<List<Line>>();
//The start (first) line, in the list of given lines, to be processed to find the first set of connected lines
Line startLineforTheNextSetOfConnectedLines=givenLines.get(0);
startLineforTheNextSetOfConnectedLines.setUsed(true);
usedCounter = 1;
//Process the list of given lines until all the lines have been used to form the connected lines
while (usedCounter < givenLines.size()) {
linesIndex = 0;
List<Line> connectedLines = new ArrayList<Line>();
connectedLines.add(linesIndex, startLineforTheNextSetOfConnectedLines);
Line nextLine=null;
//Starting with startLineforTheNextSetOfConnectedLines, the variable lastArrangedLine will hold the next lines qualifying to become the connected line
Line lastArrangedLine=startLineforTheNextSetOfConnectedLines;
//Create the list of connected lines starting with startLineforTheNextSetOfConnectedLines
for (int i = 0; i < givenLines.size(); i++) {
for (int j = 0; j < givenLines.size(); j++) {
nextLine=givenLines.get(j);
if (!nextLine.isUsed() && lastArrangedLine.getX2() == nextLine.getX1()
&& lastArrangedLine.getY2() == nextLine.getY1()) {
nextLine.setUsed(true);
usedCounter++;
connectedLines.add(++linesIndex, nextLine);
lastArrangedLine = nextLine;
break;
}
}
}
//Add the list of connected lines (found from the above nested for loops) to the list of connected lines
listOfConnectedLines.add(connectedLines);
//Find the start (first) line for the next set of connected lines
for (int i = 0; i < givenLines.size(); i++) {
if(!givenLines.get(i).isUsed()) {
startLineforTheNextSetOfConnectedLines=givenLines.get(i);
startLineforTheNextSetOfConnectedLines.setUsed(true);
usedCounter++;
break;
}
}
}
//Display the lists of connected lines
for(List<Line> connectedLines:listOfConnectedLines)
System.out.println(connectedLines);
}
}
给定行列表的输出:
[Line [x1=3350, y1=1500, x2=3200, y2=1500], Line [x1=3200, y1=1500, x2=3200, y2=1750], Line [x1=3200, y1=1750, x2=1450, y2=1750], Line [x1=1450, y1=1750, x2=1450, y2=2200], Line [x1=1450, y1=2200, x2=2100, y2=2200]]
[Line [x1=2400, y1=2200, x2=2550, y2=2200], Line [x1=2550, y1=2200, x2=2550, y2=2350], Line [x1=2550, y1=2350, x2=2850, y2=2350], Line [x1=2850, y1=2350, x2=2850, y2=2700], Line [x1=2850, y1=2700, x2=3350, y2=2700]]
[Line [x1=3650, y1=2700, x2=3750, y2=2700], Line [x1=3750, y1=2700, x2=3750, y2=2600], Line [x1=3750, y1=2600, x2=5250, y2=2600], Line [x1=5250, y1=2600, x2=5250, y2=2350], Line [x1=5250, y1=2350, x2=5000, y2=2350]]
[Line [x1=4700, y1=2350, x2=4350, y2=2350], Line [x1=4350, y1=2350, x2=4350, y2=1600], Line [x1=4350, y1=1600, x2=3650, y2=1600], Line [x1=3650, y1=1600, x2=3650, y2=1500]]
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