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:

K​i​​: h​i​​[1] h​i​​[2] ... h​i​​[K​i​​]

where K​i​​ (>0) is the number of hobbies, and h​i​​[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

一，關注點

1，關於擁有相同愛好的兩個人，如何連線？當時使用的是hobby的vector裡新增人的方式，然後兩兩結合。這是比較直白的方式。

2，標準方法不是開vector來儲存人，而是直接用陣列來儲存第一個有此愛好的人，以後的人都與這個人連線。這種方法簡單。

二，我的程式碼

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int max_n = 1100;
int N = 0;
int father[max_n];
bool flag[max_n];
int num_of_child[max_n];
vector<int> vec[max_n];
void init() {
	for (int i = 1; i <= N; i++) {
		father[i] = i;
	}
}
bool cmp(int a, int b) {
	return a > b;
}
int findFather(int x) {
	int a = x;
	while (x != father[x]) {
		x = father[x];
	}
	while (a != father[a]) {
		int z = a;
		a = father[a];
		father[z] = x;
	}
	return x;
}
void Union(int x, int y) {
	int faA = findFather(x);
	int faB = findFather(y);
	if (faA != faB) {
		father[faA] = faB;
	}
}
int main() {
	int num = 0, id = 0, num_of_circle = 0, max_id = -1;
	scanf("%d", &N);
	init();
	for (int i = 1; i <= N; i++) {
		scanf("%d:", &num);
		for (int j = 0; j < num; j++) {
			scanf("%d", &id);
			vec[id].push_back(i);
			if (id > max_id) {
				max_id = id;
			}
		}
	}
	for (int i = 0; i <= max_id; i++) {
		int temp = vec[i].size();
		for (int j = 0; j < temp - 1; j++) {
			Union(vec[i][j], vec[i][j + 1]);
		}
	}
	for (int i = 1; i <= N; i++) {
		flag[findFather(i)] = true;
		num_of_child[findFather(i)]++;
	}
	for (int i =1; i <= N; i++) {
		if (flag[i] == true) {
			num_of_circle++;
		}
	}
	sort(num_of_child + 1, num_of_child + N + 1, cmp);
	printf("%d\n", num_of_circle);
	for (int i = 1; i <= num_of_circle; i++) {
		printf("%d", num_of_child[i]);
		if (i != num_of_circle) {
			printf(" ");
		}
	}
	return 0;
}

三，標準程式碼

#include<cstdio>
#include<algorithm>
using namespace std;
const int max_n = 1010;
int father[max_n] = { 0 };
int group_num[max_n] = { 0 };
bool flag[max_n] = { false };
int course[max_n] = { 0 };
int findFather(int x) {
	int a = x;
	while (x != father[x]) {
		x = father[x];
	}
	while (a != father[a]) {
		int z = a;
		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;
	}
}
bool cmp(int a, int b) {
	return a > b;
}
int main() {
	int N = 0;
	scanf("%d", &N);
	for (int i = 1; i <= N; i++) {
		father[i] = i;
	}
	for (int i = 1; i <= N; i++) {
		int num = 0;
		scanf("%d:", &num);
		for (int j = 0; j < num; j++) {
			int hobby = 0;
			scanf("%d", &hobby);
			if (course[hobby] == 0) {
				course[hobby] = i;
			}
			Union(i, course[hobby]);
		}
	}

	for (int i = 1; i <= N; i++) {
		if (father[i] != 0) {
			flag[findFather(i)] = true;
			++group_num[findFather(i)];
		}
	}
	sort(group_num + 1, group_num + N + 1, cmp);
	int count = 0;
	for (int i = 1; i <= N; i++) {
		if (flag[i] == true)count++;
	}
	printf("%d\n", count);
	int num = 0;
	for (int i = 1; i <= N; i++) {
		if (group_num[i] != 0) {
			printf("%d", group_num[i]);
			num++;
			if (num != count) {
				printf(" ");
			}
		}
	}
	return 0;
}