「java迷宫最短路径」java走迷宫最短路径
本篇文章给大家谈谈java迷宫最短路径，以及java走迷宫最短路径对应的知识点，希望对各位有所帮助，不要忘了收藏本站喔。
本文目录一览：
- 1、Java迷宫算法问题（用栈实现）有算法简述
- 2、用java语言求最短路径
- 3、用java求最短路径问题，求源程序
- 4、迷宫数组 最短路径算法 我， 我， 我给分60！
- 5、挑战程序设计竞赛上的问题，迷宫的最短路径
Java迷宫算法问题（用栈实现）有算法简述
import java.io.File;
import java.io.FileNotFoundException;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
public class Test {
public static void main(String[] args) throws FileNotFoundException {
Scanner input = new Scanner(new File("Maze2tong.txt"));
int row = 0;
char[][] Mazemap = new char[12][58];
while (input.hasNext()) {
String line = input.nextLine();
for (int column = 0; column = line.length() - 1; column++) {
char c = line.charAt(column);
Mazemap[row][column] = c;
}
row++;
}
for (int i = 0; i 12; i++) {
for (int j = 0; j 58; j++) {
System.out.print(Mazemap[i][j]);
}
System.out.print("\n");
}
LinkedListTwoTupleInteger, Integer trace = new LinkedListTwoTupleInteger, Integer();
System.out.println(maze(Mazemap, trace));
System.out.println(trace);
}
public static boolean maze(char[][] maze,
ListTwoTupleInteger, Integer trace) {
LinkedListTwoTupleInteger, Integer path = new LinkedListTwoTupleInteger, Integer();
HashSetTwoTupleInteger, Integer traverse = new HashSetTwoTupleInteger, Integer();
for (int i = 0; i maze.length; i++) {
for (int j = 0; j maze[i].length; j++) {
if (maze[i][j] == 'S') {
path.add(new TwoTupleInteger, Integer(i, j));
}
}
}
while (!path.isEmpty()) {
TwoTupleInteger, Integer temp = path.pop();
if (traverse.contains(temp)) {
continue;
} else if (maze[temp.first][temp.second] == 'F') {
trace.add(temp);
return true;
} else if (!traverse.contains(temp)) {
if (temp.second + 1 maze[temp.first].length
maze[temp.first][temp.second + 1] != 'W')
path.add(new TwoTupleInteger, Integer(temp.first,
temp.second + 1));
if (temp.second - 1 0
maze[temp.first][temp.second - 1] != 'W')
path.add(new TwoTupleInteger, Integer(temp.first,
temp.second - 1));
if (temp.first + 1 maze.length
maze[temp.first + 1][temp.second] != 'W')
path.add(new TwoTupleInteger, Integer(temp.first + 1,
temp.second));
if (temp.first - 1 0
maze[temp.first - 1][temp.second] != 'W')
path.add(new TwoTupleInteger, Integer(temp.first - 1,
temp.second));
traverse.add(temp);
trace.add(temp);
}
}
trace.clear();
return false;
}
}
class TwoTupleA, B {
public final A first;
public final B second;
public TwoTuple(A a, B b) {
first = a;
second = b;
}
@Override
public int hashCode() {
return first.hashCode() + second.hashCode();
}
@Override
public boolean equals(Object obj) {
if (!(obj instanceof TwoTuple)) {
}
return obj instanceof TwoTuple first.equals(((TwoTuple) obj).first)
second.equals(((TwoTuple) obj).second);
}
public String toString() {
return "(" + first + ", " + second + ")";
}
} // /:-
import java.io.File;
import java.io.FileNotFoundException;
import java.util.LinkedList;
import java.util.Scanner;
class MyPoint
{
public boolean visited=false;
public int parentRow=-1;
public int parentColumn=-1;
public final char content;
public int x;
public int y;
public MyPoint(char c,int x,int y)
{
this.content = c;
this.x = x;
this.y = y;
}
}
public class Maze
{
public static MyPoint[][] getMazeArray(){
Scanner input = null;
MyPoint[][] mazemap = new MyPoint[12][58];
try {
input = new Scanner(new File("Maze2tong.txt"));
int row = 0;
while (input.hasNext()) {
String line = input.nextLine();
for (int column = 0; column = line.length() - 1; column++) {
char c = line.charAt(column);
MyPoint point = new MyPoint(c,row,column);
mazemap[row][column] = point;
}
row++;
}
input.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
return mazemap;
}
public static boolean tomRun(MyPoint[][] maze,MyPoint end)
{
int x = maze.length;
int y = maze[0].length;
LinkedListMyPoint stack=new LinkedListMyPoint();
for (int i = 0; i maze.length; i++) {
for (int j = 0; j maze[i].length; j++) {
if (maze[i][j].content == 'S') {
stack.push(maze[i][j]);
maze[i][j].visited=true;
}
}
}
boolean result=false;
while(!stack.isEmpty())
{
MyPoint t=stack.pop();
//System.out.println("pop point:"+t.x+" "+t.y+" value:"+maze[t.x][t.y]);
if(t.content == 'F')
{
result=true;
end.x = t.x;
end.y = t.y;
break;
}
if(t.x-1 0 maze[t.x-1][t.y].visited==false maze[t.x-1][t.y].content != 'W')
{
stack.push(maze[t.x-1][t.y]);
maze[t.x-1][t.y].parentRow=t.x;
maze[t.x-1][t.y].parentColumn=t.y;
maze[t.x-1][t.y].visited=true;
}
if(t.x + 1 x maze[t.x+1][t.y].visited==false maze[t.x+1][t.y].content != 'W')
{
stack.push(maze[t.x+1][t.y]);
maze[t.x+1][t.y].parentRow=t.x;
maze[t.x+1][t.y].parentColumn=t.y;
maze[t.x+1][t.y].visited=true;
}
if(t.y - 1 0 maze[t.x][t.y - 1].visited==false maze[t.x][t.y-1].content != 'W')
{
stack.push(maze[t.x][t.y-1]);
maze[t.x][t.y-1].parentRow=t.x;
maze[t.x][t.y-1].parentColumn=t.y;
maze[t.x][t.y-1].visited=true;
}
if( t.y + 1 y maze[t.x][t.y + 1].visited==false maze[t.x][t.y+1].content != 'W')
{
stack.push(maze[t.x][t.y+1]);
maze[t.x][t.y+1].parentRow=t.x;
maze[t.x][t.y+1].parentColumn=t.y;
maze[t.x][t.y+1].visited=true;
}
}
return result;
}
public static void show(int x,int y,MyPoint[][] visited)
{
if(visited[x][y].parentRow==-1)
{
System.out.println("["+x+","+y+"]");
return;
}
show(visited[x][y].parentRow,visited[x][y].parentColumn,visited);
System.out.println("-"+"["+x+","+y+"]");
}
public static void main(String[] args)
{
MyPoint[][] maze = getMazeArray();
MyPoint point = new MyPoint('c',1,1);
if(tomRun(maze,point))
{
System.out.println("逃生路径如下：");
show(point.x,point.y,maze);
}
else
System.out.println("无法走出迷宫！");
}
}
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用java语言求最短路径
最短路径就是敲代码。 这个东西行业公认,没有比敲代码学语言更加快的路了。
如果是单纯感兴趣可以买两本书自学 什么thinkinjava之类的，开始肯定看不懂的,谁开始都看不懂,摸索着来,时间长了就理解了。如果有其它语言基础学起来就快多了,因为语言这种东西,除了语法不一样,逻辑都是一样的。
如果是工作需要什么的,可以找个培训啥的。当然前提你有钱。
最后顺带吐个槽,捷径好找的话，程序员这工作就不值钱了。
用java求最短路径问题，求源程序
import java.util.Vector;
public class Link {
private Vector link = new Vector();
// private Link next = null;
public Link() {
}
public boolean addNode(Node setNode){//增加一个节点
setNode = checkNode(setNode);
if(setNode != null){
this.link.addElement((Node)setNode);
return true;
}
return false;
}
public void delNode(Node setNode){ //删除一个节点
if(!this.link.isEmpty()){
for(int i=0;i this.link.size(); i++)
{
if(setNode.getPos() == ((Node)this.link.elementAt(i)).getPos()){
this.link.remove(i);
//System.out.println("asdfasdfas:"+this.link.size());
break;
}
}
}
}
public Node checkNode(Node setNode){//判断节点是否在链表里面并取得两者的最佳值
if(!this.link.isEmpty() setNode!=null){
for(int i=0;i this.link.size(); i++)
{
if(setNode.getPos() == ((Node)this.link.elementAt(i)).getPos()){
if(setNode.getStep() ((Node)this.link.elementAt(i)).getStep()){
setNode = (Node)this.link.elementAt(i);
this.link.remove(i);
}
else
return null;
break;
}
}
}
return setNode;
}
public boolean isEmpty(){
return this.link.isEmpty();
}
public Node getBestNode(){ //得到最好的节点
Node tmpNode = null;
if(!this.link.isEmpty()){
tmpNode = (Node)this.link.elementAt(0);
//System.out.println("tmpNodeStep:"+tmpNode.getStep());
//System.out.print("OpenNode(pos,step):");
for(int i=1;i this.link.size(); i++)
{
//System.out.print("("+((Node)this.link.elementAt(i)).getPos()+","+((Node)this.link.elementAt(i)).getStep()+")");
if(tmpNode.getJudgeNum() = ((Node)this.link.elementAt(i)).getJudgeNum()){
tmpNode = (Node)this.link.elementAt(i);
}
}
}
return tmpNode;
}
}
public class FindBestPath {
private char[][] map = null;//地图
private int maxX,maxY;//最大的地图边界大小
Node startNode = null;//入口
Node endNode = null;//出口
private int endX,endY;
/*初始化
*@param setMap 地图
*@param setX,setY 边界值
//////////*@param startNode 入口
//////////*param endNode 出口
*@param sX,sY:开始点
*@param eX,eY:结束点
*/
public FindBestPath(char[][] setMap,int setX,int setY,int sX,int sY,int eX,int eY) {
this.map = setMap;
this.maxY = setX - 1; //x,y互换
this.maxX = setY - 1; //x,y互换
//this.startNode = sNode;
//this.endNode = eNode;
Node sNode = new Node();
Node eNode = new Node();
sNode.setFarther(null);
sNode.setPos(posToNum(sX,sY));
sNode.setStep(0);
eNode.setPos(posToNum(eX,eY));
this.startNode = sNode;
this.endNode = eNode;
this.endX = eX;//numToX(eNode.getPos());
this.endY = eY;//numToY(eNode.getPos());
}
public int posToNum(int x,int y){//从xy坐标获得编号
return (x+y*(this.maxY+1));
}
public int numToX(int num){//从编号获得x坐标
return (num%(this.maxY+1));
}
public int numToY(int num){//从编号获得y坐标
return (int)(num/(this.maxY+1));
}
public boolean checkVal(int x,int y){//判断是否为障碍
//System.out.println("map["+x+"]["+y+"]="+map[x][y]);
if(this.map[x][y] == 'N')
return false;
else
return true;
}
public int judge(Node nowNode){//一定要比实际距离小
//System.out.println("nowNodePos:"+nowNode.getPos());
int nowX = numToX(nowNode.getPos());
int nowY = numToY(nowNode.getPos());
int distance = Math.abs((nowX-this.endX))+Math.abs((nowY-this.endY));
// System.out.println("distance:"+distance);
return distance;
}
public Node getLeft(Node nowNode){//取得左节点
int nowX = numToX(nowNode.getPos());
int nowY = numToY(nowNode.getPos());
Node tmpNode = new Node();
if(nowY 0){//判断节点是否到最左
if(checkVal(nowX,nowY-1)){
tmpNode.setFarther(nowNode);
tmpNode.setPos(posToNum(nowX,nowY-1));
tmpNode.setStep(nowNode.getStep()+1);
tmpNode.setJudgeNum(tmpNode.getStep()+judge(tmpNode));
return tmpNode;
}
}
return null;
}
public Node getRight(Node nowNode){//取得右节点
int nowX = numToX(nowNode.getPos());
int nowY = numToY(nowNode.getPos());
Node tmpNode = new Node();
if(nowY this.maxX){//判断节点是否到最左
if(checkVal(nowX,nowY+1)){
tmpNode.setFarther(nowNode);
tmpNode.setPos(posToNum(nowX,nowY+1));
tmpNode.setStep(nowNode.getStep()+1);
tmpNode.setJudgeNum(tmpNode.getStep()+judge(tmpNode));
return tmpNode;
}
}
return null;
}
public Node getTop(Node nowNode){//取得上节点
int nowX = numToX(nowNode.getPos());
int nowY = numToY(nowNode.getPos());
Node tmpNode = new Node();
if(nowX 0){//判断节点是否到最左
if(checkVal(nowX-1,nowY)){
tmpNode.setFarther(nowNode);
tmpNode.setPos(posToNum(nowX-1,nowY));
tmpNode.setStep(nowNode.getStep()+1);
tmpNode.setJudgeNum(tmpNode.getStep()+judge(tmpNode));
return tmpNode;
}
}
return null;
}
public Node getBottom(Node nowNode){//取得下节点
int nowX = numToX(nowNode.getPos());
int nowY = numToY(nowNode.getPos());
Node tmpNode = new Node();
if(nowX this.maxY){//判断节点是否到最左
if(checkVal(nowX+1,nowY)){
tmpNode.setFarther(nowNode);
tmpNode.setPos(posToNum(nowX+1,nowY));
tmpNode.setStep(nowNode.getStep()+1);
tmpNode.setJudgeNum(tmpNode.getStep()+judge(tmpNode));
return tmpNode;
}
}
return null;
}
public Link getBestPath(){//寻找路径
Link openLink = new Link();//没有访问的路径
Link closeLink = new Link();//访问过的路径
Link path = null;//最短路径
Node bestNode = null;
Node tmpNode = null;
openLink.addNode(this.startNode);
while(!openLink.isEmpty())//openLink is not null
{
bestNode = openLink.getBestNode();//取得最好的节点
//System.out.println("bestNode:("+numToX(bestNode.getPos())+","+numToY(bestNode.getPos())+")step:"+bestNode.getJudgeNum());
if(bestNode.getPos()==this.endNode.getPos())
{
/*this.endNode.setStep(bestNode.getStep()+1);
this.endNode.setFarther(bestNode);
this.endNode.setJudgeNum(bestNode.getStep()+1);*/
path = makePath(bestNode);
break;
}
else
{
tmpNode = closeLink.checkNode(getLeft(bestNode));
if(tmpNode != null)
//System.out.println("("+numToY(tmpNode.getPos())+","+numToX(tmpNode.getPos())+")");
openLink.addNode(tmpNode);
tmpNode = closeLink.checkNode(getRight(bestNode));
if(tmpNode != null)
// System.out.println("("+numToY(tmpNode.getPos())+","+numToX(tmpNode.getPos())+")");
openLink.addNode(tmpNode);
tmpNode = closeLink.checkNode(getTop(bestNode));
if(tmpNode != null)
// System.out.println("("+numToY(tmpNode.getPos())+","+numToX(tmpNode.getPos())+")");
openLink.addNode(tmpNode);
tmpNode = closeLink.checkNode(getBottom(bestNode));
if(tmpNode != null)
// System.out.println("("+numToY(tmpNode.getPos())+","+numToX(tmpNode.getPos())+")");
openLink.addNode(tmpNode);
openLink.delNode(bestNode);
closeLink.addNode(bestNode);
}
}
return path;
}
public Link makePath(Node lastNode){//制造路径
Link tmpLink = new Link();
Node tmpNode = new Node();
int x,y;
tmpNode = lastNode;
if(tmpNode != null){
do{
x=numToX(tmpNode.getPos());
y=numToY(tmpNode.getPos());
System.out.println("map["+x+"]["+y+"]="+map[x][y]);
tmpLink.addNode(tmpNode);
tmpNode = tmpNode.getFarther();
}while(tmpNode != null);
}else
{
System.out.println("Couldn't find the path!");
}
return tmpLink;
}
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
char[][] map ={
{'Y', 'N', 'z', 'y', 'x', 'w', 'v', 'N', 'N', 'N'},
{'Y', 'N', '1', 'N', 'N', 'N', 'u', 't', 'N', 'N'},
{'N', '1', '2', '1', '1', '1', 'N', 's', 'N', 'N'},
{'N', 'N', '1', 'N', '9', 'N', 'q', 'r', 'N', 'N'},
{'N', 'N', '1', 'N', 'n', 'o', 'p', 'N', 'N', 'N'},
{'N', '4', '5', '6', 'm', 'N', 'N', 'N', 'N', 'N'},
{'N', '3', 'N', '5', 'l', 'k', 'j', 'N', 'N', 'N'},
{'N', 'N', '3', '4', 'N', 'd', 'i', 'd', 'N', 'N'},
{'N', '1', 'N', 'N', '1', 'N', 'h', 'N', 'N', 'N'},
{'N', '1', 'N', 'N', '1', 'N', 'g', 'N', 'N', 'N'},
{'N', 'a', 'b', 'c', 'd', 'e', 'f', 'N', 'N', 'N'}
};
/*map[x][y]
*如上所示：maxY=10 maxX=11 横的代表maxY,竖的代表maxX 可以自己替换
*地图的读取是
*for(i=1;i行的最大值;i++)
* for(j=1;j列的最大值;j++)
* map[i][j] = 地图[i][j]
*/
Link bestPath = new Link();
/*startNode.setFarther(null);
startNode.setPos(21);
startNode.setStep(0);
//endNode.setFarther(startNode);
endNode.setPos(79);
//endNode.setStep(0);*/
FindBestPath path = new FindBestPath(map, 11, 10, 10, 1, 0, 2);
//FindBestPath path = new FindBestPath(map, 11, 10, startNode, endNode);
bestPath = path.getBestPath();
//bestPath.printLink();
}
}
public class Node {
private int step;//从入口到该节点经历的步数
private int pos;//位置
private Node farther;//上一个结点
private int judgeNum;
public Node() {
}
public void setStep(int setStep){
this.step = setStep;
}
public int getStep(){
return this.step;
}
public void setPos(int setPos){
this.pos = setPos;
}
public int getPos(){
return this.pos;
}
public void setFarther(Node setNode){
this.farther = setNode;;
}
public Node getFarther(){
return this.farther;
}
public void setJudgeNum (int setInt){
this.judgeNum = setInt;;
}
public int getJudgeNum(){
return this.judgeNum;
}
}
迷宫数组 最短路径算法 我， 我， 我给分60！
//对角线也可以走：如从1,3到2,4最短路径为根号2
//如果对角线不想，两个for循环条件判断稍微修改下：
//代码如下：
//输出-1 表示找不到路径
/*
给个数组，0表示通路，1表示阻碍，然后给任意两个0的坐标，可以找到它们之间的最短路径并打印出来
比如说数组是 1 0 0 0 1
1 1 1 0 1
1 1 0 0 1
1 0 0 0 1
1 0 1 1 1
然后我就可以找到第一行第二个数和第五行第二个数之间的最短路径了
*/
#include stdlib.h
#include math.h
#include windows.h
#include stdio.h
#define M 5
#define N 5
typedef struct coordinate{
short int x;
short int y;
} COORDINATE;
COORDINATE path[M*N];//全局变量，存储已访问结点
float getminpath(COORDINATE *,COORDINATE *,char [][N],COORDINATE *);
void main()
{
memset(path,0,sizeof(COORDINATE)*M*N);
char maze[M][N]={
'\1','\0','\0','\0','\1',
'\1','\1','\1','\0','\1',
'\1','\1','\0','\0','\1',
'\1','\0','\0','\0','\1',
'\1','\0','\1','\1','\1',
};
COORDINATE a,b;
float dis;
do{
system("cls");
printf("请输入起点坐标：\n如1,2为第一行第二个数\n");
printf("取值范围1～%d,1～%d\n",M,N);
scanf("%hd,%hd",a.x,a.y);
printf("请输入终点坐标：\n如5,2为第五行第二个数\n");
printf("取值范围1～%d,1～%d\n",M,N);
scanf("%hd,%hd",b.x,b.y);
a.x--;
a.y--;
b.x--;
b.y--;
dis=getminpath(a,b,maze,path);
printf("最短路径为：%f\n",dis);
system("pause");
} while(getchar()!=EOF);
}
float getminpath(COORDINATE *from,COORDINATE *to,char maze[][5],COORDINATE *visited)
//从from结点到to结点可以分成两步，1.从from到from的相邻结点。2.再到to结点，这就是一个递归的过程
//将已访问结点存放入visited中
{
COORDINATE *nearpt=(COORDINATE*)malloc(sizeof(COORDINATE)*1);
memset(nearpt,0,sizeof(COORDINATE));
int i,j;
float min=0;//存储最短路径的值
COORDINATE *pvisit;//遍历已访问结点的指针
char reflag;//相邻结点是否 “已访问” 的flag
float dis=-1;//距离
if(from==0||to==0||maze==0)
{
return -1;
}
else
{
if(from-x==to-xfrom-y==to-y)//坐标相同，表示已到达
{
dis=0;
}
else
{
for(i=-1;i=1;i++)
{
for(j=-1;j=1;j++)
{
if(i==0j==0)
{
continue;
}
nearpt-x=from-x+i;
nearpt-y=from-y+j;
if(nearpt-x0||nearpt-xM-1||nearpt-y0||nearpt-yN-1)//越界
{
continue;
}
if(maze[nearpt-x][nearpt-y]!='\0')//非零为障碍，继续下一次循环
{
continue;
}
reflag='\0';//初值为'\0'，表示未访问过
for(pvisit=path;pvisitvisited;pvisit++)//搜索已访问结点
{
if(pvisit-x==nearpt-xpvisit-y==nearpt-y)
{
reflag='\1';//该结点已经访问过了
break;
}
}
if(reflag=='\1')
{
continue;
}
else
{
visited-x=from-x;//将from结点加入已访问结点数组中
visited-y=from-y;
dis=getminpath(nearpt,to,maze,visited+1);//递归
if(dis=0)
{
dis+=sqrt((nearpt-x-from-x)*(nearpt-x-from-x)+(nearpt-y-from-y)*(nearpt-y-from-y));
// printf("%d,%d到%d,%d\t距离：%3f\n",from-x,from-y,to-x,to-y,dis);
if(min==0)//第一次，还没将dis的值赋给m过
{
min=dis;
}
else
{
if(dismin)
{
min=dis;
}
else
{
}
}
}
visited-x=0;//清除from结点
visited-y=0;
}
}
}
}
}
free(nearpt);
if(min0)
{
return min;
}
else
{
return dis;
}
}
挑战程序设计竞赛上的问题，迷宫的最短路径
typedef pairint, int p;
这就是p的定义啊，p是一个类型
至于 pair 还是学一下比较好，很常用的。
pair 的定义大概是这样：
templateclass _T1, class _T2
struct pair
{
_T1 first;
_T2 second;
// ....
}
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