Chromatic judgement bipartite graph

subject ：

Input n m Express n vertices m side , There may be double edges and self rings . Judge whether the graph is a bipartite graph . 

Ideas ：

A bipartite graph is a value that divides a set of points into two halves , There are no edges in each set , But it can form an edge with points in another set , If and only if a graph does not have odd rings, it is a bipartite graph , Odd ring means that the number of sides in the ring is odd , Dyeing method ： Dye a graph with only 1 and 2 Two colors ,dfs Every point , Connected dyed in different colors , If you find that the two points are connected and the color is the same , Then there is an odd ring , It must not be a bipartite graph .

Code ： 

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/*
 Since the graph does not contain odd rings , So there must be no contradiction in the dyeing process 
 To judge a bipartite graph by coloring 

for  All points  v ：
    if v  No staining 
        dfs(v ,1)  // v Dyeing  1 Color number 
 */
public class Two_Fen_Gra {
    static int N = 1000, M = 20000, idx = 0;
    static int[] h = new int[N], e = new int[M], ne = new int[M];
    static int[] color = new int[N];// 0 It means you haven't dyed ,1,2 Represents two colors 

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] s = br.readLine().split(" ");
        int n = Integer.parseInt(s[0]), m = Integer.parseInt(s[1]);
        for (int i = 1; i <= n; i++) {
            h[i] = -1;
        }
        while (m != 0) {
            String[] s2 = br.readLine().split(" ");
            int a = Integer.parseInt(s2[0]);
            int b = Integer.parseInt(s2[1]);
            add(a, b);//  Undirected graph 
            add(b, a);
            --m;
        }
        boolean flag = true;
        for (int i = 1; i <= n; i++) {
            if (color[i] == 0) {//  If this dot has not been dyed ,
                if (!dfs(i, 1)) { //  spot  i  Dyeing failed , Then it shows that the graph is not a bipartite graph 
                    flag = false;
                    break;
                }
            }
        }

        if (flag) System.out.println("Yes");
        else System.out.println("No");

    }

    static void add(int a, int b) {
        ne[idx] = h[a];
        e[idx] = b;
        h[a] = idx++;
    }

    /*
     *  Return to the current point v Dyeing of c Color condition 
     *  Return the failure reason ：
     *     1.v The connection point of j Dyeing failed 
     *     2.v The connection point of j And v The dyeing condition is the same 
     *  If it fails, it indicates that the graph is not a bipartite graph 
     * */
    static boolean dfs(int v, int c) {
        color[v] = c;
        for (int i = h[v]; i != -1; i = ne[i]) {
            int j = e[i];
            if (color[j] == 0) {
                if (!dfs(j, 3 - c)) return false;
            } else if (color[j] == c) return false;
        }
        return true;
    }
}

Use cases ：

4 4
1 3
1 4
2 3
2 4

Output ：

Yes