广度优先算法BFS

package myalgorithm;import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
/*BFS用于记录的位置和值的结构*/
class node
{node(int xparam,int yparam,int valparam){this.x = xparam;this.y = yparam;this.value = valparam;}int x,y,value;
}
public class ShortPath {/*全局最短路径*/public int stepnum = 999;/*构建11*11的迷宫,英雄H在(1,1)的位置出发，去解救美女M(6,8)*/char[][] graph = {{'#','#','#','#','#','#','#','#','#','#','#'},{'#','H','_','_','*','_','_','*','_','_','#'},{'#','_','_','_','_','_','_','_','_','_','#'},{'#','_','*','_','_','_','*','_','_','_','#'},{'#','_','_','_','*','_','_','_','_','*','#'},{'#','_','_','_','_','_','_','*','_','*','#'},{'#','_','*','_','_','_','_','_','M','_','#'},{'#','*','_','_','*','_','_','_','_','_','#'},{'#','_','_','_','_','_','_','_','_','_','#'},{'#','_','_','_','*','_','_','_','_','_','#'},{'#','#','#','#','#','#','#','#','#','#','#'},};/*初始标记数组都为0*/public int[][] mark = new int[graph.length][graph.length];/*每一个位置有四种选择：右下左上*/public int[][] choose = {{0,1},{1,0},{0,-1},{-1,0}};/*BFS算法*/public void BFS(node startPoint){//起始点装入队列Queue<node> queue = new LinkedList<node>();startPoint.value = 1;//确保起始步数为1
        queue.offer(startPoint);node t1;int tx,ty;top:while(!queue.isEmpty()){//取队首，出队后不再入队，value也自此固定t1 = queue.poll();mark[t1.x][t1.y] = t1.value;//标记步数for(int i=0;i<4;i++){tx = t1.x + choose[i][0];ty = t1.y + choose[i][1];//找到美女，肯定是最短的，可以立即返回if(graph[tx][ty] == 'M'){stepnum = t1.value + 1;//下一步可到mark[tx][ty] = stepnum;break top;}//继续接着找，把空路径添加到队列末尾//不是炸弹和围墙,并且没有被标记if(graph[tx][ty] != '#' && graph[tx][ty] != '*'&&mark[tx][ty] == 0){queue.offer(new node(tx,ty,t1.value+1));}}}}/*BFS回溯路径*/private void getPath(node target) {int beforex = 0,beforey = 0;int x = target.x;int y = target.y;for(int i=0;i<4;i++){beforex = x + choose[i][0];beforey = y + choose[i][1];//找到英雄的出发点，函数结束if (beforex == 1 && beforey == 1){return;}if(mark[x][y] - 1 == mark[beforex][beforey]){System.out.print("("+beforex+","+beforey+")<--");break;}}getPath(new node(beforex,beforey,0));return;}
    /*main函数*/public static void main(String[] args) {ShortPath my = new ShortPath();long start = System.currentTimeMillis();my.BFS(new node(1,1,0));long end = System.currentTimeMillis();System.out.println("BFS step: " + my.stepnum + " time:" + (end-start));//打印标记结果for (int m = 0;m<my.graph.length;m++)System.out.println(Arrays.toString(my.mark[m]));//打印巡回路径my.getPath(new node(6,8,0));       }}

BFS step: 13 time:4
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 1, 2, 3, 0, 7, 8, 0, 10, 11, 0]
[0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0]
[0, 3, 0, 5, 6, 7, 0, 9, 10, 11, 0]
[0, 4, 5, 6, 0, 8, 9, 10, 11, 0, 0]
[0, 5, 6, 7, 8, 9, 10, 0, 12, 0, 0]
[0, 6, 0, 8, 9, 10, 11, 0, 13, 0, 0]
[0, 0, 10, 9, 0, 11, 0, 0, 0, 0, 0]
[0, 0, 11, 10, 11, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 11, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
(5,8)<--(4,8)<--(4,7)<--(4,6)<--(4,5)<--(3,5)<--(3,4)<--(3,3)<--(2,3)<--(2,2)<--(2,1)

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扩展：

PhotoShop中的魔术棒选择工具的原理，就是从鼠标选中的点作为种子，并加入队列；依次查找周边的点，如果颜色与种子颜色相近，则加入队列；对队列中的元素依次进行类似扩展，这样就选取了与最初种子颜色相同的点的集合。