What is binary enumeration ？
link : Binary enumeration .

Seven segment code

link : Seven segment code .

This question is a little circuitous , The code has been modified for a while

There are a little more judgment conditions here .

The whole idea is ：

• Enumerate all the situations
• For each case, judge whether it is connected by searching the set

Enumeration, I use binary enumeration here , The advantage is that it is relatively fast , But here is a question to fill in the blank , There are only 120 What kind of situation , use DFS It's OK, too

Judgment of connectivity is mainly divided into and and Judge Two aspects

• and ： As long as the two lights are connected, they merge .

I think too much here , Because we are Before and after , Without repeating judgment , Don't think about breaking or because it has two ends , Head inconsistency .

• Judge ： After we've sorted out the search collection , as long as pre[t] Or myself (t) The situation is a connected section （ You can draw pictures to understand ）. in other words , The number of segments is greater than 1, It's not connected

Complete code

import java.util.Scanner;
public class test {
    
	
	public static int count=0;
	public static boolean[][] l = new boolean[8][8];
	public static void init() {
    
		l[1][1] = l[1][2]=l[1][6] = true;
		l[2][2] = l[2][1]=l[2][7] = l[2][3]=true;
		l[3][3] = l[3][2]=l[3][4] = l[3][7]=true;
		l[4][4] = l[4][3]=l[4][5] = true;
		l[5][5] = l[5][4]=l[5][7] = l[5][6]=true;
		l[6][6] = l[6][1]=l[6][7] = l[6][5]=true;
		l[7][7] = l[7][2]=l[7][3] = l[7][5]=l[7][6]=true;
	}
	public static void main(String[] args) {
    
		init();
		for(int i=0;i<(1<<7);i++) {
    
			int[] light = new int[8];
			for(int j=0;j<7;j++) {
    
				if((i&(1<<j))>0) {
    
					light[j+1]=1;
				}
			}
			System.out.println();
			if(check(light)) {
    
				count++;
			}
			
		}
		System.out.println(count);
	}
	
	public static boolean check(int[] lights) {
    
		int i=1,j=1;
		int[] pre= {
    0,1,2,3,4,5,6,7};
		while(i<8) {
    
			if(lights[i]==1) {
    
				for(j=i+1;j<8;j++) {
    
					if(lights[j]==1) {
    
						
						if(l[i][j]) {
    
							join(i,j,pre);
						}
					}
				}
			}
			i++;
		}
		int count=0;
		for(int t=1;t<8;t++) {
    
			if(lights[t]==1&&pre[t]==t)
				count++;
		}
		return count==1;
	}
	
	public static int find(int x,int[] pre) {
    
		if(pre[x] == x) return x;
		return pre[x] = find(pre[x],pre);
	}
	
	public static void join(int x,int y,int[] pre) {
    
		int px = find(x,pre),py = find(y,pre);
		if(px!=py)
			pre[px] = py;
	}
	
	
	
 

}