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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 (≤), 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

    参照柳婼大佬的并查集解法:https://www.liuchuo.net/archives/2183

    #include <cstdio>
    #include <vector>
    #include <algorithm>
    using namespace std;
    vector<int> father, isRoot;
    int cmp1(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 a, int b) {
        int faA = findFather(a);
        int faB = findFather(b);
        if(faA != faB) father[faA] = faB;
    }
    int main() {
        int n, k, t, cnt = 0;
        int course[1001] = {0};
        scanf("%d", &n);
        father.resize(n + 1);
        isRoot.resize(n + 1);
        for(int i = 1; i <= n; i++)
            father[i] = i;
        for(int i = 1; i <= n; i++) {
            scanf("%d:", &k);
            for(int j = 0; j < k; j++) {
                scanf("%d", &t);
                if(course[t] == 0)
                    course[t] = i;
                Union(i, findFather(course[t]));
            }
        }
        for(int i = 1; i <= n; i++)
            isRoot[findFather(i)]++;
        for(int i = 1; i <= n; i++) {
            if(isRoot[i] != 0) cnt++;
        }
        printf("%d
    ", cnt);
        sort(isRoot.begin(), isRoot.end(), cmp1);
        for(int i = 0; i < cnt; i++) {
            printf("%d", isRoot[i]);
            if(i != cnt - 1) printf(" ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/littlepage/p/12875586.html
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