PAT甲级1107

Posted 2022-08-30 wx630c98f24f6b8

1107. Social Clusters (30)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

/*
#include<cstdio>
#include<vector>
#include<set>
#include<algorithm>
using namespace std;
const int maxn = 1000;
bool hashtable[maxn] =  false ;
set<int> person[maxn+1];
vector<int> clusters;
bool isCommon(set<int> s1, set<int> s2)

  set<int>::iterator it;
  bool flag = false;
  for (it = s2.begin(); it != s2.end(); it++)
  
    if (s1.find(*(it)) != s1.end())
    
      flag = true;
      break;
    
  
  return flag;

void unionset(set<int> &s1, set<int> s2)

  for (set<int>::iterator it = s2.begin(); it != s2.end(); it++)
  
    s1.insert(*(it));
  

int cmp(set<int> s1, set<int> s2)

  return s1.size() > s2.size();

int main()

  int N;
  scanf("%d", &N);
  int k, h;
  for (int i = 1; i <= N; i++)
  
    scanf("%d", &k);
    getchar();
    for (int j = 0; j < k; j++)
    
      scanf("%d", &h);
      person[i].insert(h);
    
  
  sort(person+1, person + 1+N, cmp);
  vector<int> v;
  int sum = 0;
  for (int i = 1; i <= N; i++)
  
    int count = 1;
    if (!hashtable[i])
    
      for (int j = i + 1; j <= N; j++)
      
        if (isCommon(person[i], person[j]) && !hashtable[j])
        
          count++;
          unionset(person[i], person[j]);
          hashtable[j] = true;
        
      
      for (int j = i + 1; j <= N; j++)
      
        if (isCommon(person[i], person[j]) && !hashtable[j])
        
          count++;
          unionset(person[i], person[j]);
          hashtable[j] = true;
        
      
      for (int j = i + 1; j <= N; j++)
      
        if (isCommon(person[i], person[j]) && !hashtable[j])
        
          count++;
          unionset(person[i], person[j]);
          hashtable[j] = true;
        
      
      for (int j = i + 1; j <= N; j++)
      
        if (isCommon(person[i], person[j]) && !hashtable[j])
        
          count++;
          unionset(person[i], person[j]);
          hashtable[j] = true;
        
      
      sum += count;
      if (sum <= N)
      
      v.push_back(count);
      hashtable[i] = true;
      
    
  
  printf("%d\\n", v.size());
  sort(v.begin(), v.end());
  for (int i = v.size()-1; i>=0; i--)
  
    printf("%d", v[i]);
    if (i)
      printf(" ");
    else
      printf("\\n");
  
  return 0;


本人的暴力解法，也可以全过，中间为什么要弄4个for循环，因为在顺序遍历时当set加入其它元素后，也会使得之前扫描过的满足共同兴趣
所以需要再扫描一遍
3
5
3 5
这就是活生生的例子，3跟5不是共同兴趣，而3跟3 5有共同兴趣，合并后使得5也可以并入其中
而有些情况扫描两遍还不够
*/
//下面是并查集解法
#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 1010;
int father[N];//存放父亲结点
int isRoot[N] =  0 ;//记录每个结点是否作为某个集合的根结点
int course[N] =  0 ;
/*int findFather(int x)//查找x所在集合的根结点

  int a = x;
  while (x != father[x])
  
    x = father[x];
  
  //路径压缩
  while (a != father[a])
  
    int z = a;
    a = father[a];
    father[z] = x;
  
  return x;
*/
/*
递归版本，路径压缩：*/
int findFather(int x)

  if (x == father[x])return x;
  else
  
    int F = findFather(father[x]);
    father[x] = F;
    return F;
  

/**/
void Union(int a, int b)//合并a和b所在的集合

  int faA = findFather(a);
  int faB = findFather(b);
  if (faA != faB)
    father[faA] = faB;

void init(int n)//初始化father[i]为i，且flag[i]为false

  for (int i = 1; i <= n; i++)
  
    father[i] = i;
    isRoot[i] = false;
  

bool cmp(int a, int b)//isRoot数组从大到小排序

  return a > b;

int main()

  int n, k, h;
  scanf("%d", &n);//人数
  init(n);//要记得初始化
  for (int i = 1; i <= n; i++)//对每个人
  
    scanf("%d:",&k);//活动个数
    for (int j = 0; j < k; j++)//对每个活动
    
      scanf("%d", &h);//输入i号人喜欢的活动h
      if (course[h] == 0)//如果活动h第一次有人喜欢
      
        course[h] = i;//令i喜欢活动h
      
      Union(i, findFather(course[h]));//合并
    
    
  
  for (int i = 1; i <= n; i++)
  
    isRoot[findFather(i)]++;//i的根结点是findFather(i),人数加1
  
  int ans = 0;//记录集合数目
  for (int i = 1; i <= n; i++)
  
    if (isRoot[i] != 0)
    
      ans++;//只统计isRoot[i]不为0的
    
  
  printf("%d\\n", ans);//输出集合个数
  sort(isRoot + 1, isRoot + 1 + n, cmp);//从大到小排序
  for (int i = 1; i <= ans; i++)//依次输出每个集合内的个数
  
    printf("%d", isRoot[i]);
    if (i < ans)printf(" ");
  
  return 0;