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 = ;
int course[] = {};
scanf("%d", &n);
father.resize(n + );
isRoot.resize(n + );
for(int i = ; i <= n; i++)
father[i] = i;
for(int i = ; i <= n; i++) {
scanf("%d:", &k);
for(int j = ; j < k; j++) {
scanf("%d", &t);
if(course[t] == )
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 = ; 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 = ; i <= n; i++) {
if(isRoot[i] != ) cnt++;
}
printf("%d\n", cnt);
sort(isRoot.begin(), isRoot.end(), cmp1);
for(int i = ; i < cnt; i++) {
printf("%d", isRoot[i]);
if(i != cnt - ) printf(" ");
}
return ;
}
//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。
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