When register on a social network, you are always asked to specify your hobbies in order to find some potential friends with the same hobbies. A “social cluster” is a set of people who have some of their hobbies in common. You are supposed to find all the clusters.
Input Specification:
Each input file contains one test case. For each test case, the first line contains a positive integer N (<=1000), the total number of people in a social network. Hence the people are numbered from 1 to N. Then N lines follow, each gives the hobby list of a person in the format:
Ki: hi[1] hi[2] … hi[Ki]
where Ki (>0) is the number of hobbies, and hi[j] is the index of the j-th hobby, which is an integer in [1, 1000].
Output Specification:
For each case, print in one line the total number of clusters in the network. Then in the second line, print the numbers of people in the clusters in non-increasing order. The numbers must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:
8
3: 2 7 10
1: 4
2: 5 3
1: 4
1: 3
1: 4
4: 6 8 1 5
1: 4
Sample Output:
3
4 3 1
题目大意：有n个人，每个人喜欢k个活动，如果两个人有任意一个活动相同，就称为他们处于同一个社交网络。求这n个人一共形成了多少个社交网络。
分析：并查集。先写好init、findFather、Union。
0. 每个社交圈的结点号是人的编号，而不是课程。课程是用来判断是否处在一个社交圈的。
1. course[t]表示任意一个喜欢t活动的人的编号。如果当前的课程t，之前并没有人喜欢过，那么就course[t] = i，i为它自己的编号，表示i为喜欢course[t]的一个人的编号
2. course[t]是喜欢t活动的人的编号，那么findFather(course[t])就是喜欢这个活动的人所处的社交圈子的根结点，合并根结点和当前人的编号的结点i。即Union(i, findFather(course[t]))，把它们处在同一个社交圈子里面
3. isRoot[i]表示编号i的人是不是它自己社交圈子的根结点，如果等于0表示不是根结点，如果不等于0，每次标记isRoot[findFather(i)]++，那么isRoot保存的就是如果当前是根结点，那么这个社交圈里面的总人数
4. isRoot中不为0的编号的个数cnt就是社交圈圈子的个数
5. 把isRoot从大到小排列，输出前cnt个，就是社交圈人数的从大到小的输出顺序
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#include <cstdio> #include <vector> #include <algorithm> using namespace std; vector<int> father, isRoot; int cmp1(int a, int b){return a > b;} int findFather(int x) { int a = x; while(x != father[x]) x = father[x]; while(a != father[a]) { int z = a; a = father[a]; father[z] = x; } return x; } void Union(int a, int b) { int faA = findFather(a); int faB = findFather(b); if(faA != faB) father[faA] = faB; } int main() { int n, k, t, cnt = 0; int course[1001] = {0}; scanf("%d", &n); father.resize(n + 1); isRoot.resize(n + 1); for(int i = 1; i <= n; i++) father[i] = i; for(int i = 1; i <= n; i++) { scanf("%d:", &k); for(int j = 0; j < k; j++) { scanf("%d", &t); if(course[t] == 0) course[t] = i; Union(i, findFather(course[t])); } } for(int i = 1; i <= n; i++) isRoot[findFather(i)]++; for(int i = 1; i <= n; i++) { if(isRoot[i] != 0) cnt++; } printf("%d\n", cnt); sort(isRoot.begin(), isRoot.end(), cmp1); for(int i = 0; i < cnt; i++) { printf("%d", isRoot[i]); if(i != cnt - 1) printf(" "); } return 0; } |
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