PAT 1107 Social Clusters
1107 Social Clusters
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总结： 每个人有不同的爱好，需要找到共同爱好的人组成一个社团
代码：这里发现未对map的键进行赋值，但是对其进行了访问，最后用 auto 循环的时候会打印这种情况
#include
#include
#include 柳神代码：
#include
#include
#include
using namespace std;
vector father, isRoot;
int cmp1(int a, int b){
return a > b;
}
int findFather(int x) {
int a = x;
while(x != father[x])//这是找根节点的过程
x = father[x];
while(a != father[a]) {//这里的过程是路径压缩
int z = a;
a = father[a];//这句代码和上面的循环是一样的，一步一步查找根节点
father[z] = x;//因为再上一个循环中计算出来了跟节点，而x存着根节点的值，这里将路径上的每一个点都指向了根节点
}
//while(x!=father[x]) x=findFather(father[x]);//这是递归查找父节点，也实现了路径压缩
return x;
}
void Union(int a, int b) {
int faA = findFather(a);
int faB = findFather(b);
if(faA != faB)//如果两个人不在同一个集合中，那么就加入这个集合
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);
//判断当前这个爱好是否有人， 如果之前有人有这个爱好，那么会存储着
//第一个人的id，通过并查集找到这个人，并将他加入这个集合当中
if(course[t] == 0)
course[t] = i;
Union(i, findFather(course[t]));
}
}
for(int i = 1; i <= n; i++)
isRoot[findFather(i)]++;//计算每个并查集里面的人数
for(int i = 1; i <= n; i++) {
if(isRoot[i] != 0)//统计并查集的个数
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;
} 好好学习，天天向上！
我要考研！