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.
题目大意 每个人都有不同的爱好，只要有一个爱好相同，那么这两个人就属于同一个社交圈，问给定的人中一共有多少个社交圈，以及每个社交圈的人数有多少分析 可以采用并查集来做。 开一个爱好数组，数组大小为1000，并将爱好数组初始为0，表示没有任何人有这个爱好，数组中存储的是第一次有此爱好的人的序号。如果第i此人的爱好数组中的值不为0，则将i和爱好数组中的人的序号的根节点合并。代码实现 #include <cstdio> #include <algorithm> using namespace std; const int N = 1001; int father[N]; int isroot[N]; int course[N]; //爱好数组 int findfather(int x){ //寻找根节点 int a = x; while(x != father[x]) x = father[x]; while(a != father[a]){ //路径压缩 int z = a; a = father[a]; father[z] = x; } return x; } void Union(int a, int b){ //合并两个集合 int faA = findfather(a); int faB = findfather(b); if(faA != faB){ father[faA] = faB; } } //初始化 void init(int n){ for(int i = 1; i <= n; i++){ father[i] = i; isroot[i] = 0; } for(int i = 0; i < N; i++) course[i] = 0; } bool cmp(int a, int b){ //从大到小排序 return a > b; } int main() { int n, num, a; scanf("%d", &n); init(n); //初始化 for(int i = 1; i <= n; i++){ //int num; scanf("%d:", &num); for(int j = 0; j < num; j++){ scanf("%d", &a); if(course[a] == 0) //如果爱好a第一次有人喜欢 course[a] = i; //令i喜欢活动h Union(i, findfather(course[a])); //合并 } } for(int i = 1; i <= n; i++){ //i的根节点是findfather(i),人数加1 isroot[findfather(i)]++; } //记录集合个数 int ans = 0; for(int i = 1; i <= n; i++){ if(isroot[i] != 0) ans++; } printf("%d\n", ans); sort(isroot + 1, isroot + n + 1, cmp); //从大到小排序 for(int i = 1; i <= ans; i++){ if(i != 1) printf(" %d",isroot[i]); else printf("%d", isroot[1]); } return 0; }