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
一，关注点
1，关于拥有相同爱好的两个人，如何连接？当时使用的是hobby的vector里添加人的方式，然后两两结合。这是比较直白的方式。
2，标准方法不是开vector来存储人，而是直接用数组来存储第一个有此爱好的人，以后的人都与这个人连接。这种方法简单。
二，我的代码
#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，标准代码
#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;
}