13518219792
class Base
公司专注于为企业提供成都网站设计、成都网站制作、微信公众号开发、电子商务商城网站建设，微信小程序开发，软件按需开发网站等一站式互联网企业服务。凭借多年丰富的经验，我们会仔细了解各客户的需求而做出多方面的分析、设计、整合，为客户设计出具风格及创意性的商业解决方案，创新互联更提供一系列网站制作和网站推广的服务。
{
protected String name;
protected getScore(){return 0;};
}
class Team extends Base
{
private int gf;
private int ga;
private int pts;
private int gd;
public Team(int Gf,int Ga)
{
gf=Gf;
ga=Ga;
gd=gf-ga;
if(gd0)
{
pts=3
}
else if(gd==0)
{
pts=1;
}
else
{
pts=0;
}
}
public int getScore()
{
return pts*10 + gd*5 + gf;
}
}
//大概是这样了,看在我大晚上给你写代码的份上，给我分吧
通过遗传算法走迷宫。虽然图1和图2均成功走出迷宫，但是图1比图2的路径长的多，且复杂，遗传算法可以计算出有多少种可能性，并选择其中最简洁的作为运算结果。
示例图1：
示例图2：
实现代码：
import java.util.ArrayList;
import java.util.Collections;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Random;
/**
* 用遗传算法走迷宫
*
* @author Orisun
*
*/
public class GA {
int gene_len; // 基因长度
int chrom_len; // 染色体长度
int population; // 种群大小
double cross_ratio; // 交叉率
double muta_ratio; // 变异率
int iter_limit; // 最多进化的代数
Listboolean[] individuals; // 存储当代种群的染色体
Labyrinth labyrinth;
int width; //迷宫一行有多少个格子
int height; //迷宫有多少行
public class BI {
double fitness;
boolean[] indv;
public BI(double f, boolean[] ind) {
fitness = f;
indv = ind;
}
public double getFitness() {
return fitness;
}
public boolean[] getIndv() {
return indv;
}
}
ListBI best_individual; // 存储每一代中最优秀的个体
public GA(Labyrinth labyrinth) {
this.labyrinth=labyrinth;
this.width = labyrinth.map[0].length;
this.height = labyrinth.map.length;
chrom_len = 4 * (width+height);
gene_len = 2;
population = 20;
cross_ratio = 0.83;
muta_ratio = 0.002;
iter_limit = 300;
individuals = new ArrayListboolean[](population);
best_individual = new ArrayListBI(iter_limit);
}
public int getWidth() {
return width;
}
public void setWidth(int width) {
this.width = width;
}
public double getCross_ratio() {
return cross_ratio;
}
public ListBI getBest_individual() {
return best_individual;
}
public Labyrinth getLabyrinth() {
return labyrinth;
}
public void setLabyrinth(Labyrinth labyrinth) {
this.labyrinth = labyrinth;
}
public void setChrom_len(int chrom_len) {
this.chrom_len = chrom_len;
}
public void setPopulation(int population) {
this.population = population;
}
public void setCross_ratio(double cross_ratio) {
this.cross_ratio = cross_ratio;
}
public void setMuta_ratio(double muta_ratio) {
this.muta_ratio = muta_ratio;
}
public void setIter_limit(int iter_limit) {
this.iter_limit = iter_limit;
}
// 初始化种群
public void initPopulation() {
Random r = new Random(System.currentTimeMillis());
for (int i = 0; i population; i++) {
int len = gene_len * chrom_len;
boolean[] ind = new boolean[len];
for (int j = 0; j len; j++)
ind[j] = r.nextBoolean();
individuals.add(ind);
}
}
// 交叉
public void cross(boolean[] arr1, boolean[] arr2) {
Random r = new Random(System.currentTimeMillis());
int length = arr1.length;
int slice = 0;
do {
slice = r.nextInt(length);
} while (slice == 0);
if (slice length / 2) {
for (int i = 0; i slice; i++) {
boolean tmp = arr1[i];
arr1[i] = arr2[i];
arr2[i] = tmp;
}
} else {
for (int i = slice; i length; i++) {
boolean tmp = arr1[i];
arr1[i] = arr2[i];
arr2[i] = tmp;
}
}
}
// 变异
public void mutation(boolean[] individual) {
int length = individual.length;
Random r = new Random(System.currentTimeMillis());
individual[r.nextInt(length)] ^= false;
}
// 轮盘法选择下一代,并返回当代最高的适应度值
public double selection() {
boolean[][] next_generation = new boolean[population][]; // 下一代
int length = gene_len * chrom_len;
for (int i = 0; i population; i++)
next_generation[i] = new boolean[length];
double[] cumulation = new double[population];
int best_index = 0;
double max_fitness = getFitness(individuals.get(best_index));
cumulation[0] = max_fitness;
for (int i = 1; i population; i++) {
double fit = getFitness(individuals.get(i));
cumulation[i] = cumulation[i - 1] + fit;
// 寻找当代的最优个体
if (fit max_fitness) {
best_index = i;
max_fitness = fit;
}
}
Random rand = new Random(System.currentTimeMillis());
for (int i = 0; i population; i++)
next_generation[i] = individuals.get(findByHalf(cumulation,
rand.nextDouble() * cumulation[population - 1]));
// 把当代的最优个体及其适应度放到best_individual中
BI bi = new BI(max_fitness, individuals.get(best_index));
// printPath(individuals.get(best_index));
//System.out.println(max_fitness);
best_individual.add(bi);
// 新一代作为当前代
for (int i = 0; i population; i++)
individuals.set(i, next_generation[i]);
return max_fitness;
}
// 折半查找
public int findByHalf(double[] arr, double find) {
if (find 0 || find == 0 || find arr[arr.length - 1])
return -1;
int min = 0;
int max = arr.length - 1;
int medium = min;
do {
if (medium == (min + max) / 2)
break;
medium = (min + max) / 2;
if (arr[medium] find)
min = medium;
else if (arr[medium] find)
max = medium;
else
return medium;
} while (min max);
return max;
}
// 计算适应度
public double getFitness(boolean[] individual) {
int length = individual.length;
// 记录当前的位置,入口点是（1,0）
int x = 1;
int y = 0;
// 根据染色体中基因的指导向前走
for (int i = 0; i length; i++) {
boolean b1 = individual[i];
boolean b2 = individual[++i];
// 00向左走
if (b1 == false b2 == false) {
if (x 0 labyrinth.map[y][x - 1] == true) {
x--;
}
}
// 01向右走
else if (b1 == false b2 == true) {
if (x + 1 width labyrinth.map[y][x + 1] == true) {
x++;
}
}
// 10向上走
else if (b1 == true b2 == false) {
if (y 0 labyrinth.map[y - 1][x] == true) {
y--;
}
}
// 11向下走
else if (b1 == true b2 == true) {
if (y + 1 height labyrinth.map[y + 1][x] == true) {
y++;
}
}
}
int n = Math.abs(x - labyrinth.x_end) + Math.abs(y -labyrinth.y_end) + 1;
// if(n==1)
// printPath(individual);
return 1.0 / n;
}
// 运行遗传算法
public boolean run() {
// 初始化种群
initPopulation();
Random rand = new Random(System.currentTimeMillis());
boolean success = false;
while (iter_limit-- 0) {
// 打乱种群的顺序
Collections.shuffle(individuals);
for (int i = 0; i population - 1; i += 2) {
// 交叉
if (rand.nextDouble() cross_ratio) {
cross(individuals.get(i), individuals.get(i + 1));
}
// 变异
if (rand.nextDouble() muta_ratio) {
mutation(individuals.get(i));
}
}
// 种群更替
if (selection() == 1) {
success = true;
break;
}
}
return success;
}
// public static void main(String[] args) {
// GA ga = new GA(8, 8);
// if (!ga.run()) {
// System.out.println("没有找到走出迷宫的路径.");
// } else {
// int gen = ga.best_individual.size();
// boolean[] individual = ga.best_individual.get(gen - 1).indv;
// System.out.println(ga.getPath(individual));
// }
// }
// 根据染色体打印走法
public String getPath(boolean[] individual) {
int length = individual.length;
int x = 1;
int y = 0;
LinkedListString stack=new LinkedListString();
for (int i = 0; i length; i++) {
boolean b1 = individual[i];
boolean b2 = individual[++i];
if (b1 == false b2 == false) {
if (x 0 labyrinth.map[y][x - 1] == true) {
x--;
if(!stack.isEmpty() stack.peek()=="右")
stack.poll();
else
stack.push("左");
}
} else if (b1 == false b2 == true) {
if (x + 1 width labyrinth.map[y][x + 1] == true) {
x++;
if(!stack.isEmpty() stack.peek()=="左")
stack.poll();
else
stack.push("右");
}
} else if (b1 == true b2 == false) {
if (y 0 labyrinth.map[y - 1][x] == true) {
y--;
if(!stack.isEmpty() stack.peek()=="下")
stack.poll();
else
stack.push("上");
}
} else if (b1 == true b2 == true) {
if (y + 1 height labyrinth.map[y + 1][x] == true) {
y++;
if(!stack.isEmpty() stack.peek()=="上")
stack.poll();
else
stack.push("下");
}
}
}
StringBuilder sb=new StringBuilder(length/4);
IteratorString iter=stack.descendingIterator();
while(iter.hasNext())
sb.append(iter.next());
return sb.toString();
}
}
代码实现(Java)
1. 输入
(1) 代表地图二值二维数组(0表示可通路，1表示路障)
int[][] maps = {
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0 },
{ 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0 },
{ 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 }
};123456789123456789
(2) 按照二维数组的特点，坐标原点在左上角，所以y是高，x是宽，y向下递增，x向右递增，我们将x和y封装成一个类，好传参，重写equals方法比较坐标(x,y)是不是同一个。
public class Coord
{
public int x;
public int y;
public Coord(int x, int y)
{
this.x = x;
this.y = y;
}
@Override
public boolean equals(Object obj)
{
if (obj == null) return false;
if (obj instanceof Coord)
{
Coord c = (Coord) obj;
return x == c.x y == c.y;
}
return false;
}
}12345678910111213141516171819202122231234567891011121314151617181920212223
(3) 封装路径结点类，字段包括：坐标、G值、F值、父结点，实现Comparable接口，方便优先队列排序。
public class Node implements Comparable
{
public Coord coord; // 坐标
public Node parent; // 父结点
public int G; // G：是个准确的值，是起点到当前结点的代价
public int H; // H：是个估值，当前结点到目的结点的估计代价
public Node(int x, int y)
{
this.coord = new Coord(x, y);
}
public Node(Coord coord, Node parent, int g, int h)
{
this.coord = coord;
this.parent = parent;
G = g;
H = h;
}
@Override
public int compareTo(Node o)
{
if (o == null) return -1;
if (G + H o.G + o.H)
return 1;
else if (G + H o.G + o.H) return -1;
return 0;
}
}1234567891011121314151617181920212223242526272829303112345678910111213141516171819202122232425262728293031
(4) 最后一个数据结构是A星算法输入的所有数据，封装在一起，传参方便。:grin:
public class MapInfo
{
public int[][] maps; // 二维数组的地图
public int width; // 地图的宽
public int hight; // 地图的高
public Node start; // 起始结点
public Node end; // 最终结点
public MapInfo(int[][] maps, int width, int hight, Node start, Node end)
{
this.maps = maps;
this.width = width;
this.hight = hight;
this.start = start;
this.end = end;
}
}12345678910111213141516171234567891011121314151617
2. 处理
(1) 在算法里需要定义几个常量来确定：二维数组中哪个值表示障碍物、二维数组中绘制路径的代表值、计算G值需要的横纵移动代价和斜移动代价。
public final static int BAR = 1; // 障碍值
public final static int PATH = 2; // 路径
public final static int DIRECT_VALUE = 10; // 横竖移动代价
public final static int OBLIQUE_VALUE = 14; // 斜移动代价12341234
(2) 定义两个辅助表：Open表和Close表。Open表的使用是需要取最小值，在这里我们使用Java工具包中的优先队列PriorityQueue，Close只是用来保存结点，没其他特殊用途，就用ArrayList。
Queue openList = new PriorityQueue(); // 优先队列(升序)
List closeList = new ArrayList();1212
(3) 定义几个布尔判断方法：最终结点的判断、结点能否加入open表的判断、结点是否在Close表中的判断。
/**
* 判断结点是否是最终结点
*/
private boolean isEndNode(Coord end,Coord coord)
{
return coord != null end.equals(coord);
}
/**
* 判断结点能否放入Open列表
*/
private boolean canAddNodeToOpen(MapInfo mapInfo,int x, int y)
{
// 是否在地图中
if (x 0 || x = mapInfo.width || y 0 || y = mapInfo.hight) return false;
// 判断是否是不可通过的结点
if (mapInfo.maps[y][x] == BAR) return false;
// 判断结点是否存在close表
if (isCoordInClose(x, y)) return false;
return true;
}
/**
* 判断坐标是否在close表中
*/
private boolean isCoordInClose(Coord coord)
{
return coord!=nullisCoordInClose(coord.x, coord.y);
}
/**
* 判断坐标是否在close表中
*/
private boolean isCoordInClose(int x, int y)
{
if (closeList.isEmpty()) return false;
for (Node node : closeList)
{
if (node.coord.x == x node.coord.y == y)
{
return true;
}
}
return false;
}1234567891011121314151617181920212223242526272829303132333435363738394041424344454612345678910111213141516171819202122232425262728293031323334353637383940414243444546
(4) 计算H值，“曼哈顿” 法，坐标分别取差值相加
private int calcH(Coord end,Coord coord)
{
return Math.abs(end.x - coord.x) + Math.abs(end.y - coord.y);
}12341234
(5) 从Open列表中查找结点
private Node findNodeInOpen(Coord coord)
{
if (coord == null || openList.isEmpty()) return null;
for (Node node : openList)
{
if (node.coord.equals(coord))
{
return node;
}
}
return null;
}123456789101112123456789101112
(6) 添加邻结点到Open表
/**
* 添加所有邻结点到open表
*/
private void addNeighborNodeInOpen(MapInfo mapInfo,Node current)
{
int x = current.coord.x;
int y = current.coord.y;
// 左
addNeighborNodeInOpen(mapInfo,current, x - 1, y, DIRECT_VALUE);
// 上
addNeighborNodeInOpen(mapInfo,current, x, y - 1, DIRECT_VALUE);
// 右
addNeighborNodeInOpen(mapInfo,current, x + 1, y, DIRECT_VALUE);
// 下
addNeighborNodeInOpen(mapInfo,current, x, y + 1, DIRECT_VALUE);
// 左上
addNeighborNodeInOpen(mapInfo,current, x - 1, y - 1, OBLIQUE_VALUE);
// 右上
addNeighborNodeInOpen(mapInfo,current, x + 1, y - 1, OBLIQUE_VALUE);
// 右下
addNeighborNodeInOpen(mapInfo,current, x + 1, y + 1, OBLIQUE_VALUE);
// 左下
addNeighborNodeInOpen(mapInfo,current, x - 1, y + 1, OBLIQUE_VALUE);
}
/**
* 添加一个邻结点到open表
*/
private void addNeighborNodeInOpen(MapInfo mapInfo,Node current, int x, int y, int value)
{
if (canAddNodeToOpen(mapInfo,x, y))
{
Node end=mapInfo.end;
Coord coord = new Coord(x, y);
int G = current.G + value; // 计算邻结点的G值
Node child = findNodeInOpen(coord);
if (child == null)
{
int H=calcH(end.coord,coord); // 计算H值
if(isEndNode(end.coord,coord))
{
child=end;
child.parent=current;
child.G=G;
child.H=H;
}
else
{
child = new Node(coord, current, G, H);
}
openList.add(child);
}
else if (child.G G)
{
child.G = G;
child.parent = current;
// 重新调整堆
openList.add(child);
}
}
}1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606112345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061
(7) 回溯法绘制路径
private void drawPath(int[][] maps, Node end)
{
if(end==null||maps==null) return;
System.out.println("总代价：" + end.G);
while (end != null)
{
Coord c = end.coord;
maps[c.y][c.x] = PATH;
end = end.parent;
}
}12345678910111234567891011
(8) 开始算法，循环移动结点寻找路径，设定循环结束条件，Open表为空或者最终结点在Close表
public void start(MapInfo mapInfo)
{
if(mapInfo==null) return;
// clean
openList.clear();
closeList.clear();
// 开始搜索
openList.add(mapInfo.start);
moveNodes(mapInfo);
}
/**
* 移动当前结点
*/
private void moveNodes(MapInfo mapInfo)
{
while (!openList.isEmpty())
{
if (isCoordInClose(mapInfo.end.coord))
{
drawPath(mapInfo.maps, mapInfo.end);
break;
}
Node current = openList.poll();
closeList.add(current);
addNeighborNodeInOpen(mapInfo,current);
}
}
单元和区域和数值，，，中的最大
