connect components in undirected graph

Description

输入一个简单无向图，求出图中连通块的数目。

Input

输入的第一行包含两个整数n和m，n是图的顶点数，m是边数。1<=n<=1000，0<=m<=10000。

以下m行，每行是一个数对v y，表示存在边(v,y)。顶点编号从1开始。
Output

单独一行输出连通块的数目。

Sample Input
 Copy sample input to clipboard
5 3
1 2
1 3
2 4
Sample Output
2

View Code 
 1 #include <iostream>
 2 using namespace std;
 3 
 4 #define max 1001
 5 
 6 int rank[max];
 7 int father[max];
 8 
 9 void ini(int n)
10 {
11     for (int i = 1; i <= n; i++){
12         father[i] = i;
13         rank[i] = 1;
14     }
15 }
16 
17 int find(int x)
18 {
19     if (father[x] != x)
20        father[x] = find(father[x]);
21     return father[x];
22 }
23 
24 void Union(int x, int y)
25 {
26     int fatherX = find(x);
27     int fatherY = find(y);
28     if (fatherX == fatherY)
29         return;
30     if (rank[fatherX] >= rank[fatherY]){
31         father[fatherX] = fatherY;
32         rank[fatherX] += rank[fatherY];
33     }
34     else{
35          father[fatherY] = fatherX;
36          rank[fatherY] += rank[fatherX];     
37     }
38 }
39 
40 int main()
41 {
42     int n, m, a, b;
43     cin >> n >> m;
44     ini(n);
45     for (int i = 0; i < m; i++){
46         cin >> a >> b;
47         Union(a, b);
48     }
49     int count = 0;
50     for (int i = 1; i <= n; i++){
51         if (father[i] == i){
52            count++;
53            //cout << i << " + " << endl;
54         }
55     }
56     cout << count << endl;
57  
58    // system("PAUSE");
59     return 0;
60 }

 

posted @ 2012-12-22 13:32  gumcstronger  阅读(328)  评论(0)    收藏  举报