gpt4 book ai didi

java - Knight's Tour GUI 处理中

转载 作者:行者123 更新时间:2023-11-30 03:38:33 26 4
gpt4 key购买 nike

我正在使用基本 gui 解决骑士之旅问题,我想在两个文本字段中获取用户输入,这两个文本字段组成了用户的 (x,y),然后在一个文本框中打印,如果解决方案可行的话,并且在另一个中,我写了骑士采用的路径。我的算法工作正常,但我在 gui 中遇到问题。我给了 (x,y) 一些默认值,以便我得到正确的输出。但是当我更改值时(x,y) 在文本字段中,没有发生任何变化。这是主文件,下面还有另一个事件处理程序文件。我们将衷心感谢您的帮助。我正在处理 2.2.1。这就是如何输出屏幕看起来像 enter image description here

主文件/project.pde

// Need G4P library
import g4p_controls.*;
Maxim maxim;
AudioPlayer player;
int x=-1;
int y=-1;
String solution="";
PImage img;
int count=0;

public void setup(){
size(480, 320, JAVA2D);
maxim=new Maxim(this);
player=maxim.loadFile("song.wav");
player.setLooping(true);

img=loadImage("chess.jpg");

createGUI();
customGUI();

// Place your setup code here

}
int t[]=new int[25];
boolean visited[][]=new boolean[5][5];
int check[][]=new int[5][5];
boolean b;
int counter=-1;
boolean move(int x,int y , int m){
boolean result=false;
if (x<0 || x>=5 || y<0 || y>=5 || visited[x][y]==true)
{
return false;
}

visited[x][y]=true;

if (m==24)
{

visited[x][y]=true;

return true;
}
else
{
String xstring=String.valueOf(x);
String ystring=String.valueOf(y);
solution=solution+xstring+","+ystring+" ";
print (x);
print(",");
print(y);
check[x][y]=counter+1;
if (move(x+2,y+1,m+1) || move(x+2,y-1,m+1)
|| move(x-2,y+1,m+1) || move(x-2,y-1,m+1)
|| move(x+1,y+1,m+1) || move(x+1,y-1,m+1)
|| move(x-1,y+1,m+1) || move(x-1,y-1,m+1)){
print (x);
print(",");
print(y);
//check[x][y]=1;
return true;

}


return false;


}
}


public void draw(){
counter=counter+1;
background(0,128,128);
image(img,0,0,480,320);
player.play();
textarea2.setText(solution);


String txt1 = textfield1.getText();
x = Integer.parseInt(txt1);

String txt2 = textfield2.getText();
y= Integer.parseInt(txt2);
print(solution);
if(x>=0 && y>=0)
{
b=move(x,y,0);


if(b==false)
{
textarea1.setText("Solution is not possible,enter other coordinates");
}
if(b==true)
{
textarea1.setText("Congratulations solution is possible");
}
}
if(count%8==0)
{
delay(1000);
println(counter);

}

}
void keyPressed()
{
if (key==13)
{
solution="";
print(solution);
textarea2.setText(solution);
String txt1 = textfield1.getText();
x = Integer.parseInt(txt1);

String txt2 = textfield2.getText();
y= Integer.parseInt(txt2);
}
if(x>=0 && y>=0)
{
b=move(x,y,0);


if(b==false)
{
textarea1.setText("Solution is not possible,enter other coordinates");
}
if(b==true)
{
textarea1.setText("Congratulations solution is possible");
}
}
}

// Use this method to add additional statements
// to customise the GUI controls
public void customGUI(){

}

这是事件处理程序文件

/* =========================================================
* ==== WARNING ===
* =========================================================
* The code in this tab has been generated from the GUI form
* designer and care should be taken when editing this file.
* Only add/edit code inside the event handlers i.e. only
* use lines between the matching comment tags. e.g.

void myBtnEvents(GButton button) { //_CODE_:button1:12356:
// It is safe to enter your event code here
} //_CODE_:button1:12356:

* Do not rename this tab!
* =========================================================
*/

public void tf1(GTextField source, GEvent event) { //_CODE_:textfield1:418637:
println("textfield1 - GTextField >> GEvent." + event + " @ " + millis());
} //_CODE_:textfield1:418637:

public void tf2(GTextField source, GEvent event) { //_CODE_:textfield2:859413:
println("textfield2 - GTextField >> GEvent." + event + " @ " + millis());
} //_CODE_:textfield2:859413:

public void ta1(GTextArea source, GEvent event) { //_CODE_:textarea1:252891:
println("textarea1 - GTextArea >> GEvent." + event + " @ " + millis());
} //_CODE_:textarea1:252891:

public void ta2(GTextArea source, GEvent event) { //_CODE_:textarea2:483845:
println("textarea2 - GTextArea >> GEvent." + event + " @ " + millis());
} //_CODE_:textarea2:483845:

public void slider1_change1(GSlider source, GEvent event) { //_CODE_:slider1:280049:
println("slider1 - GSlider >> GEvent." + event + " @ " + millis());
} //_CODE_:slider1:280049:

public void slider2_change1(GSlider source, GEvent event) { //_CODE_:slider2:362722:
println("slider2 - GSlider >> GEvent." + event + " @ " + millis());
} //_CODE_:slider2:362722:



// Create all the GUI controls.
// autogenerated do not edit
public void createGUI(){
G4P.messagesEnabled(false);
G4P.setGlobalColorScheme(GCScheme.BLUE_SCHEME);
G4P.setCursor(ARROW);
if(frame != null)
frame.setTitle("Sketch Window");
textfield1 = new GTextField(this, 210, 32, 160, 30, G4P.SCROLLBARS_NONE);
textfield1.setText("1");
textfield1.setPromptText("Enter x-Cordinate");
textfield1.setOpaque(true);
textfield1.addEventHandler(this, "tf1");
textfield2 = new GTextField(this, 204, 96, 160, 30, G4P.SCROLLBARS_NONE);
textfield2.setText("1");
textfield2.setPromptText("Enter y Cordinate");
textfield2.setLocalColorScheme(GCScheme.PURPLE_SCHEME);
textfield2.setOpaque(true);
textfield2.addEventHandler(this, "tf2");
textarea1 = new GTextArea(this, 53, 196, 160, 80, G4P.SCROLLBARS_NONE);
textarea1.setLocalColorScheme(GCScheme.GREEN_SCHEME);
textarea1.setOpaque(true);
textarea1.addEventHandler(this, "ta1");
textarea2 = new GTextArea(this, 288, 192, 160, 80, G4P.SCROLLBARS_NONE);
textarea2.setLocalColorScheme(GCScheme.YELLOW_SCHEME);
textarea2.setOpaque(true);
textarea2.addEventHandler(this, "ta2");
slider1 = new GSlider(this, 96, 276, 264, 40, 10.0);
slider1.setLimits(0.5, 0.0, 1.0);
slider1.setNumberFormat(G4P.DECIMAL, 2);
slider1.setOpaque(false);
slider1.addEventHandler(this, "slider1_change1");
slider2 = new GSlider(this, 348, 240, 100, 36, 10.0);
slider2.setLimits(0.5, 0.0, 1.0);
slider2.setNumberFormat(G4P.DECIMAL, 2);
slider2.setOpaque(false);
slider2.addEventHandler(this, "slider2_change1");
}

// Variable declarations
// autogenerated do not edit
GTextField textfield1;
GTextField textfield2;
GTextArea textarea1;
GTextArea textarea2;
GSlider slider1;
GSlider slider2;

最佳答案

上述结构存在一些问题。您在 draw() 方法中执行递归 move() 方法。但在处理中,draw() 每秒被动画线程调用多次。

此类情况的常见设计是:

  • 您应该有保存应用程序状态(逻辑状态)的变量
  • 您的 draw() 方法绘制的内容仅取决于状态
  • 状态可以被动画线程或任何其他线程修改

我建议稍微更改一下您的代码:

首先 - 状态变量:

// 1 - display status, wait to enter values
// 2 - check if x and y are correct
// 3 - solve problem
int state=1;
boolean solutionExists=false;

public void setup() {
...
}

下一个draw()方法:

void draw() {

counter=counter+1;
background(0,128,128);

String coords = "(" + x + "," + y + ")";

if (state == 1) {
if(solutionExists) {
textarea1.setText("Congratulations solution is possible " + coords);
} else {
textarea1.setText("Solution is not possible,enter other coordinates " +coords);
}

return;
}

if (state == 2) {
readXY();
return;
}

if (state == 3) {
println("find solution for: " + coords);
solutionExists = move(x,y,0);
state = 1;
return;
}

}

public void readXY() {
try {
x = Integer.parseInt(textfield1.getText().trim());
y = Integer.parseInt(textfield2.getText().trim());
state = 3;
} catch(Exception e) {
state = 1;
}

}

最后是文本字段处理程序:

public void tf1(GTextField source, GEvent event) {
if (event.getType().equals("LOST_FOCUS")) {
state=2;
}
}

public void tf2(GTextField source, GEvent event) {
if (event.getType().equals("LOST_FOCUS")) {
state=2;
}
}

如您所见:

  • if state==1 - draw() 仅更新消息
  • if state==2 - draw() 检查 x 和 y 是否有效,如果有效 -> 将状态更改为 3
  • if state==3 - draw() 执行递归算法,更新solutionExists 变量,将状态更改为1

只要您的文本字段 1 或文本字段 2 失去焦点,它就会将状态更改为 2。

  • draw() 仅由应用程序状态驱动。
  • 应用程序状态会被其他事件修改。

为了获得最佳结果,递归算法应该在另一个线程中执行,而不是在动画线程中。

一个注意事项:当您编辑文本字段时,它可能包含诸如 "" (空)或 "3" (前导空格)或 "3 之类的字符串“ 等 - 此类文本无法使用 Integer.parseInt 进行解析 - 您需要修剪此类文本并确保不会抛出 NumberFormatException - 请参阅 readXY() 方法。

关于java - Knight's Tour GUI 处理中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27296494/

26 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com