题目链接:https://pintia.cn/problem-sets/994805342720868352/problems/994805361586847744
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
题目类型:并查集。
大概翻译:有n个人,每人喜欢k个活动,若两人喜欢的活动存在相同的一项则他们属于同一个社交网络,求一共形成了多少社交网络。
思路:先将getRoot()、Merge()函数写好,getRoot函数用于获取祖先节点,Merge函数用于压缩路径。在每次读如具体的喜好时都将先寻找所在树,进行并树;最后统计isRoot数组内不为0的个数即为一共多少棵树(isRoot内每个不为0的数即为每棵树所含的节点数即有多少个人在同一个社交网络),将isRoot排序输出。
源代码:
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
vector<int> parent, isRoot; //父母数组、树的个数
int course[1010]; //course[t]表示任意⼀一个喜欢t活动的⼈人的编号;
int cmp(int a, int b) {return a > b;}
int getRoot(int x) { //得到祖先节点
int a = x;
while(x != parent[x])
x = parent[x];
//下面一段参考与柳婼的代码,用于优化时间,否则会有测试用例超时
while(a != parent[a]) {
int z = a;
a = parent[a];
parent[z] = x;
}
return x;
}
void Merge(int a, int b) { //合并a, b所属的树
int pa = getRoot(a);
int pb = getRoot(b);
if (pa != pb) parent[pa] = pb; //将a树并到b树下面。
}
int main() {
int n, k, t, cnt = 0;
scanf("%d", &n); //n个人
parent.resize(n + 1); //resize函数用于重置vector数组的大小
isRoot.resize(n + 1);
//在输入之前,每个节点都可以看成祖先是自己的节点
for (int i = 1; i <= n; i++) parent[i] = i;
for (int i = 1; i <= n; i++) {
scanf("%d:", &k); //所喜欢的k个活动
for (int j = 0; j < k; j++) {
scanf("%d", &t); //具体活动
//如果当前的课程t,之前并没有⼈人喜欢过,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。
if (course[t] == 0) course[t] = i;
Merge(i, getRoot(course[t])); //得到合并后的树
}
}
//统计树的个数,即为一共分为多少网络
for (int i = 1; i <= n; i++) isRoot[getRoot(i)]++;
for(int i = 1; i <= n; i++) if(isRoot[i] != 0) cnt++;
printf("%d\n", cnt);
//对已经统计好的数据排序,sort函数属于algorithm库
sort(isRoot.begin(), isRoot.end(), cmp);
for(int i = 0; i < cnt; i++) {
printf("%d", isRoot[i]);
if(i != cnt - 1) printf(" ");
}
return 0;
}
(代码部份思路参照柳神写出,在这膜一下!)