有6個(gè)英文字母,a,b,c ,d,e,f
?從中任意取N個(gè)(N<=6)來排列.
已知 a 只能與 b 相連,?
???????? b不能和e,f相連
?????????c不能和a相連
???????? d不能和 a ,e相連
???????????f 不能和?a,b相連
請(qǐng)打印出字母f或b 在末尾的組合順序,用Java實(shí)現(xiàn)

我的解答:

??? import java.util.*;

/**
?*
?* @author ShenXiaoliang
?*
?*/
public class DemoGraph {
?
?final static int VERTICE=6;
?
?private ArrayList<String> patheSet=new ArrayList<String>();
?
?private String[] ver={"a","b","c","d","e","f"};
?
?private int[][] graph=new int[6][6];
?
?private String path="";
?
?private boolean[] isVisit=new boolean[VERTICE];
?
?
?public DemoGraph() {
??
??initiGraph();
??for(int index=0;index<VERTICE;index++)
??depthSearch(index);
??show();
??
?}
?
?private void initiGraph(){
??graph[0][1]=1;
??graph[1][0]=1;
??graph[1][2]=1;
??graph[1][3]=1;
??graph[2][1]=1;
??graph[2][3]=1;
??graph[2][4]=1;
??graph[2][5]=1;
??graph[3][1]=1;
??graph[3][2]=1;
??graph[3][5]=1;
??graph[4][2]=1;
??graph[4][5]=1;
??graph[5][2]=1;
??graph[5][3]=1;
??graph[5][4]=1;
?}
?
?private void depthSearch(int start){
??
??isVisit[start]=true;
??path+=ver[start];
??if(path.charAt(path.length()-1)=='f'||path.charAt(path.length()-1)=='b') patheSet.add(path);
??for(int index=0;index<VERTICE;index++)
???if(graph[start][index]==1&&isVisit[index]==false)
????depthSearch(index);
???else continue;
??path=path.substring(0,path.length()-1);
??isVisit[start]=false;
?}
?
?
?private void show(){
??for(String pa:patheSet)
???System.out.println(pa);
?}

?/**
? * @param args
? */
?public static void main(String[] args) {
??// TODO Auto-generated method stub
??new DemoGraph();
?}

}

?????????