pat-a 1107. Social Clusters (30)

并查集，我写的多了一点。

简单说下就行，两个人如果有共同的爱好就是一个圈子里面的。开始错误原因在于压缩路径不够彻底。因为最后判断有几个圈子时没有去寻找根直接判断父节点，因此要压缩到底才能直接判断。

#include<cstdio>#include<algorithm>#include<functional>using namespace std;int num[1010][1010];int p[1010];int visit[1010];void init(){for(int i=0;i<1010;++i) p[i]=i;}int find(int x){if(p[x]==x) return x;else return p[x]=find(p[x]);}void merge(int i,int j){int x=find(i);int y=find(j);if(x!=y){if(x>y) p[x]=y;else p[y]=x;}}int main(){int n,t,a,ans=0,flag=0,k=0;int c[1010];scanf("%d",&n);init();for(int i=0;i<n;++i){scanf("%d:",&t);while(t--){scanf("%d",&a);num[i][a]=1;for(int j=0;j<i;++j){if(num[j][a]){merge(i,j);}}}}for(int i=0;i<n;++i){for(int j=0;j<i;++j){if(find(i)==find(j)) merge(i,j);}}//for(int i=0;i<n;++i) printf("%d ",p[i]);for(int i=0;i<n;++i){visit[p[i]]++;if(visit[p[i]]==1) ans++;}printf("%d\n",ans);for(int i=0;i<n;++i){if(visit[i]){c[k++]=visit[i];}}sort(c,c+k,greater<int>());for(int i=0;i<k;++i){if(flag)printf(" %d",c[i]);else{flag=1;printf("%d",c[i]);}}return 0;} 

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:

83: 2 7 101: 42: 5 31: 41: 31: 44: 6 8 1 51: 4

Sample Output:

34 3 1