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  • 1107 Social Clusters (复杂并查集)

    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 (≤), 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​​ (>) is the number of hobbies, and [ 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
    

    思路:

    1. 在每次输入个人的一个爱好后,都可能对如今的社交网络产生变化,故都需要在此时进行合并操作
    2. father数组在初始保存的根节点都是自己,表示自己是一个单独的社交网络
    3. 这里因为每个人可能有不止一个爱好,故还需辅助数组hobby,记录任何一个有该爱好的人,之后再与当前读入的人合并根节点即可
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    const int maxn=1010;
    int father[maxn]={0};
    int isRoot[maxn]={0};
    int hobby[maxn]={0};
    int n,num,temp;
    
    int findFather(int x){
        if(father[x]==x) return x;
        else{
            father[x]=findFather(father[x]);
            return father[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(){
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            father[i]=i;
        for(int i=1;i<=n;i++){
            scanf("%d:",&num);
            for(int j=0;j<num;j++){
                scanf("%d",&temp);
                if(hobby[temp]==0)
                    hobby[temp]=i;
                Union(i,hobby[temp]);
            }
        }
        for(int i=1;i<=n;i++){
            isRoot[findFather(i)]++;//此处不能用father[i],可能存在嵌套关系 
        }
        int ans=0;
        for(int i=1;i<=n;i++)
            if(isRoot[i]!=0) ans++;
        printf("%d
    ",ans);
        sort(isRoot+1,isRoot+n+1,cmp);
        for(int i=1;i<=ans;i++){
            printf("%d",isRoot[i]);
            if(i<ans) printf(" ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/exciting/p/10426140.html
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