Pat(A) 1107. Social Clusters (30)

原题目：

原题链接：https://www.patest.cn/contests/pat-a-practise/1107

1107. Social Clusters (30)

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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

题目大意

每个人有多个爱好，拥有相同爱好的人为同一簇。问有多少簇人，按人数有大到小输出每个簇的人数。

解题报告

并查集的题目，将拥有同一爱好的人归为同一集合。
ans[i]表示以i为根的一簇的人数，ans[0]表示总的簇数。
pre[i]表示上一个有爱好i的人的编号。

代码

#include "iostream"#include "vector"#include "algorithm"using namespace std;int father[1001] = {0};int pre[1001] = {0}; // pre[i] means the last person who has a hobby iint N;vector<int> ans;int getF(int x){    int a = x,t;    while(x != father[x]){        x = father[x];    }    while(a != father[a]){        t = a;        a = father[a];        father[t] = x;    }    return x;}void Union(int a,int b){    int fa = getF(a);    int fb = getF(b);    father[fb] = fa;}void init(){    scanf("%d",&N);    int i,k,j,x;    ans.resize(N + 1,0);    for(i = 0; i < 1001; i++)        father[i] = i;    for(i = 1; i <= N; i++){        scanf("%d: ",&k);        for(j = 0; j < k; j ++){            scanf("%d",&x);            if(pre[x] == 0)                pre[x] = i;            Union(i,getF(pre[x]));        }    }}void cal(){    int  i,f;    for(i = 1; i <= N; i++){        f = getF(i);        if( f == i)            ans[0]++;        ans[getF(i)] ++;    }    sort(ans.begin() + 1,ans.end());}int main(){    init();    cal();    printf("%d\n",ans[0]);    if(ans[0]){        printf("%d",ans[ans.size() - 1]);        for(int i = ans.size() - 2,k = 1; k < ans[0]; i--,k++)            printf(" %d",ans[i]);        printf("\n");    }    system("pause");;}