FIG determined staining bipartite

Address https://www.acwing.com/problem/content/862/

Given a free edges m n points to view the drawing there may be multiple edges and loopback.

Please determine whether this figure is a bipartite graph.

Input Format

The first line contains two integers n and m.

Subsequently m rows, each row comprising two integers u and v, indicates the presence of points u and v between an edge.

Output Format

If a given graph is a bipartite graph, the output is "Yes", otherwise a "No".

data range

Sample input:

Sample output:

Yes

The basic template

FIG traverse point is then randomly assigned adjacent to the color dots of another color must be

Template code

 1 #include <cstring>
 2 #include <iostream>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 const int N = 100010, M = 200010;
 8 
 9 int n, m;
10 int h[N], e[M], ne[M], idx;
11 int color[N];
12 
13 void add(int a, int b)
14 {
15     e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
16 }
17 
18 bool dfs(int u, int c)
19 {
20     color[u] = c;
21 
22     for (int i = h[u]; i != -1; i = ne[i])
23     {
24         int j = e[i];
25         if (!color[j])
26         {
27             if (!dfs(j, 3 - c)) return false;
28         }
29         else if (color[j] == c) return false;
30     }
31 
32     return true;
33 }
34 
35 int main()
36 {
37     scanf("%d%d", &n, &m);
38 
39     memset(h, -1, sizeof h);
40 
41     while (m -- )
42     {
43         int a, b;
44         scanf("%d%d", &a, &b);
45         add(a, b), add(b, a);
46     }
47 
48     bool flag = true;
49     for (int i = 1; i <= n; i ++ )
50         if (!color[i])
51         {
52             if (!dfs(i, 1))
53             {
54                 flag = false;
55                 break;
56             }
57         }
58 
59     if (flag) puts("Yes");
60     else puts("No");
61 
62     return 0;
63 }

View Code

Own Code

. 1 #include <the iostream>
 2 #include <Vector>
 . 3 #include <algorithm>
 . 4 #include < String >
 . 5  
. 6  
. 7  the using  namespace STD;
 . 8  
. 9  / * 
10  Input format
 11  The first line contains two integers n and m.
12 is  
13 is  the next m rows, each row comprising two integers u and v, indicates the presence of points u and v between an edge.
14  
15  Output Format
 16  if a given graph is a bipartite graph, the output "Yes", and otherwise outputs "No".
. 17  
18 is  data range
 . 19  1 ≦ n, m≤105
 20 is  input sample:
 21 is  . 7. 7
 22 is  . 1. 3
 23 is  . 1. 4
 24 2. 3
 25  2. 4
 26 is  . 5. 6
 27  . 6. 7
 28  . 5. 7
 29  sample output:
 30  No
 31 is  * / 
32  
33 is  const  int N = 100010 , M = 200010 ;
 34 is  
35  int n-, m;
 36  
37 [ Vector <Vector < int UV >> ( 100010 );
 38 is  
39 Vector < int > ColorPt ( 100010 );
 40  
41 is  String RET;
 42 is  
43 is  void DFS (int IDX, int Color) {
 44 is      IF (IDX> n-RET || == " No " ) {
 45          return ;
 46 is      }
 47  
48      ColorPt [IDX] = Color;
 49      
50      // color 
51 is      Color Color% = 2 + . 1 ;
 52 is      for ( int I = 0 ; I <UV [IDX] .size (); I ++ ) {
 53 is          // adjacent to the point dyed another color point 
54 is          int coloridx = UV [IDX] [I];
 55          //Coloring conflicts directly returns 
56 is          IF (ColorPt [coloridx] =! 0 && ColorPt [coloridx] =! Color)
 57 is          {
 58              RET = " No " ;
 59              return ;
 60          }
 61 is          IF (ret.empty () && ColorPt [coloridx ] == 0 )
 62 is              DFS (coloridx, Color);
 63 is      }
 64  
65  
66      return ;
 67  }
 68  
69  
70  int main ()
 71 is  {
 72      CIN n->> >> m;
73 
74     for (int i = 0; i < m; i++) {
75         int a; int b;
76         cin >> a >> b;
77         uv[a].push_back(b);
78         uv[b].push_back(a);
79     }
80     /*for(int i= 1;i < n;i++)
81         dfs(i,1);*/
82 
83     for (int i = 1; i <= n; i++) {
84         if (ret == "No") break;
85         if (colorr[i] != 0)  continue;
86         dfs(i, 1);
87     }
88 
89     if (ret.empty())  ret = "Yes";
90     cout << ret << endl;
91 
92 
93     return 0;
94 }

View Code