Jobdu-1526-朋友圈[解题代码] | Acm之家


九度-1526-朋友圈[解题代码] | Acm之家
假如已知有n个人和m对好友关系(存于数字r)。如果两个人是直接或间接的好友(好友的好友的好友…),则认为他们属于同一个朋友圈,请写程序求出这n个人里一共有多少个朋友圈。
假如:n = 5 , m = 3 , r = {{1 , 2} , {2 , 3} , {4 , 5}},表示有5个人,1和2是好友,2和3是好友,4和5是好友,则1、2、3属于一个朋友圈,4、5属于另一个朋友圈,结果为2个朋友圈。
输入:
输入包含多个测试用例,每个测试用例的第一行包含两个正整数 n、m,1=<n,m<=100000。接下来有m行,每行分别输入两个人的编号f,t(1=<f,t<=n),表示f和t是好友。 当n为0时,输入结束,该用例不被处理。
输出:
对应每个测试用例,输出在这n个人里一共有多少个朋友圈。

  1.     public static void main(String[] args) throws Exception{  
  2.         StreamTokenizer st = new StreamTokenizer(System.in);  
  3.         while (st.nextToken() != StreamTokenizer.TT_EOF) {  
  4.             int n = (int) st.nval;  
  5.             if (n == 0) {  
  6.                 break;  
  7.             }  
  8.             if (n == 1) {  
  9.                 System.out.println(1);  
  10.             }else {  
  11.                 int []parent = new int[n+1];  
  12.                 for (int i = 1; i <= n; i++) {  
  13.                     parent[i] = i;  
  14.                 }  
  15.                 st.nextToken() ;  
  16.                 int m = (int) st.nval;  
  17.                 for (int i = 0; i < m; i++) {  
  18.                     st.nextToken() ;  
  19.                     int f = (int) st.nval;  
  20.                     st.nextToken() ;  
  21.                     int t = (int) st.nval;  
  22.                     union(f ,t , parent );  
  23.                 }  
  24.                    
  25.                 for (int i = 1; i < n+1; i++) {  
  26.                     parent[i] = findParent(i, parent);  
  27.                 }  
  28.                 Set<Integer> numSet = new HashSet<Integer>();  
  29.                 for (int i = 1; i < n+1; i++) {  
  30.                     numSet.add(parent[i]);  
  31.                 }  
  32.                 System.out.println( numSet.size());  
  33.             }  
  34.         }  
  35.     }  
  36.     
  37.     private static void union(int f, int t, int[] parent ) {  
  38.             
  39.         int a = findParent(f , parent);  
  40.         int b = findParent(t , parent);  
  41.         if (a == b) return;   
  42.         if (a > b) {     
  43.             parent[a] = b;     
  44.          } else {  
  45.             parent[b] = a;   
  46.          }  
  47.     }  
  48.     
  49.     private static int findParent(int f, int[] parent) {  
  50.         if (parent[f] == f) {  
  51.             return f;  
  52.         }  
  53.         return findParent(parent[f],parent );  
  54.     }

  1. private static int findParent(int f) {  
  2.     while (parent[f] != f) {  
  3.         f = parent[f];  
  4.     }  
  5.     return f;  
http://www.cnblogs.com/xinsheng/p/3581057.html
int find(int x) {
    if(father[x] != x) {
        father[x] = find(father[x]);
    }
    return father[x];
}
void merge(int a, int b) {
    int fa = find(a);
    int fb = find(b);
    if(fa == fb)
        return;
    father[fa] = fb;
}
int main() {
    int n, m;
    while(scanf("%d", &n) && n != 0) {
        scanf("%d", &m);
        for(int i = 1; i <= n; i ++)
            father[i] = i;

        int a, b;
        for(int i = 0; i < m; i ++) {
            scanf("%d%d", &a, &b);
            merge(a,b);
        }
        int cnt = 0;
        for(int i = 1; i <= n; i ++) {
            if(father[i] == i)
                cnt++;
        }
        cout << cnt << endl;
    }
    return 0;
}
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