PAT-1107 Social Clusters (30 分) 并查集模板

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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[*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

题目大意：有n个人，每个人喜欢k个活动，如果两个人有任意一个活动相同，就称为他们处于同一个社交网络。求这n个人一共形成了多少个社交网络。

第一个喜欢这个活动的称为教主,之后喜欢这个活动的人和教主unite即可.

1.并查集写错了,返回一个等号写成了两个等号.

2.找每组属于谁的时候可能还没有压缩路径,所以应该统计find(i)而不是par[i],,找分为几组的时候才可以用par[i]==i

#include <iostream>
#include<bits/stdc++.h>
#define each(a,b,c) for(int a=b;a<=c;a++)
#define de(x) cout<<#x<<" "<<(x)<<endl
using namespace std;

const int maxn=1000+5;
const int inf=0x3f3f3f3f;

int par[maxn];
int Rank[maxn];
int jiaozhu[maxn];
multiset<int>MS;
void init(int n)
{
    for(int i=1;i<=n;i++)///初始化小错误
    {
        par[i]=i;
        Rank[i]=0;
    }
}
int find(int x)
{
    return par[x]==x?x:par[x]=find(par[x]);///并查集的父节点满足par[x]==x,如果不是的话也要把par[x]更新,返回的是find(par[x])同时进行了路径的压缩
}
priority_queue<int>Q;
void unite(int x,int y)
{
    //de(x);
    //de(y);
    x=find(x);
    y=find(y);

    if(x==y)return;
    if(Rank[x]<Rank[y])
    {
        par[x]=y;///军衔小,认y当爹
    }
    else
    {
        par[y]=x;
        if(Rank[x]==Rank[y])Rank[x]++;///本来是兄弟,x却当了爹
    }
}
int root_cnt[maxn];
/**
8
3: 2 7 10
1: 4
2: 5 3
1: 4
1: 3
1: 4
4: 6 8 1 5
1: 4
*/
int ans[maxn];
int cur=0;
int cmp(int a,int b){return a>b;}
int main()
{
    int n;
    scanf("%d",&n);
    init(n);
    cur=0;
    each(i,1,n)
    {
        int k;
        scanf("%d: ",&k);
        while(k--)
        {
            int temp;
            scanf("%d",&temp);
            if(jiaozhu[temp]==0)
            {
                jiaozhu[temp]=i;
            }
            else
            {
                unite(jiaozhu[temp],i);

            }
        }
    }
    each(i,1,n)
    {
        root_cnt[find(i)]++;
    }
    for(int i=1;i<=n;i++)
    {
        if(root_cnt[i]!=0)
        {
            ans[cur++]=root_cnt[i];
        }
    }
    sort(ans,ans+cur,cmp);
    printf("%d\n",cur);
    each(i,0,cur-1)
    {
        printf("%d",ans[i]);
        printf(i==cur-1?"\n":" ");
    }
}

PAT-1107 Social Clusters (30 分) 并查集模板

标签：HERE   name   contain   cat   ota   index   The   and   应该   

原文地址：https://www.cnblogs.com/Tony100K/p/11773848.html