1107 Social Clusters[并查集][难]
1107 Social Clusters(30 分)
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:根据输入将用户分堆,也就是并查集的题目;输入格式:一共几个人,第几个人一共有几个爱好:爱好序号。根据兴趣爱好进行分堆,看哪些人在一起了,并按人数非升序排列。
//题目中给的样例共有3个簇,最大的4个人包括 2,4,6,8四个人,3个人的包括3,5,7三个人,这种没有重复,但是有没有可能会重复呢?
//我能想到的就是创建1000个向量,每个用户分别push,然后求出来长度!=1的向量,然后按照长度大小输出,其他没什么想法了。。。
代码来自:https://www.liuchuo.net/archives/2183
#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])//先找到整棵树的根节点。 x = father[x]; while(a != father[a]) {//找父亲的时候顺便将其高度变短,父节点都变为根节点了。 int z = a;//你学过的!变成两层高的。 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;//也就是将这个课程归并为i所有。 Union(i, findFather(course[t]));//找到这个课程的父亲,因为两者有相同的父亲, //此处应该findFather(course[t])是父亲的,因为其是course的编号。 //是一个簇内的,那么合并即可。 } } for(int i = 1; i <= n; i++) isRoot[findFather(i)]++;//那些是根节点的都++;最终求出来的是簇内所含的人数。 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; }
//真的太神奇了。
1.couser[i]表示爱好i第一次出现时的用户编号,如果=0表示这个爱好没出现。
2.如果新来了一个人k爱好也为i,那么将couser[i]和Kunion即可。
3.复习了并查集,并查集三要素:findFather+union+初始化father数组为自己;
4.这里使用了一个for循环遍历来确定每个簇有多少个人。