PAT甲组 1107. Social Clusters (30)

重温一下前几天的一个题，放松一下放松一下，今天大家都太强了QAQ

1107. Social Clusters (30)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题意：给个N，有N个人，接下来按编号1~N的顺序，每行给出一个人的感兴趣的活动，先给个K，表示K个活动，然后引号后面有K个活动编号，表示这个人喜欢这个活动。如果有两个人有任意重叠的喜欢的活动，它们就是一个小圈子的，要求输出有几个小圈子，接下来降序输出每个小圈子有几个人。

思路：并查集大法（千万不要一边读一边合并，不知道为啥一边读一边合并有三个测试点就是过不去，惆怅）↓

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
const int maxv = 1111;

int fa[maxv],root[maxv];

int findfa(int x)//找爸爸
{
int a=x;
while(x!=fa[x])
{
x=fa[x];
}
while(a!=fa[a])
{
int z=a;
a=fa[a];
fa[z]=x;
}
return x;
}

void uni(int a,int b)//合并两个小朋友
{
int faa=findfa(a);
int fab=findfa(b);
if(faa!=fab)
{
fa[faa]=fab;
}
}

struct node
{
int num;//喜欢该活动的人数
int a[maxv];//喜欢该活动的人的编号
}act[maxv];

int N,K,temp;

int main()
{
scanf("%d",&N);
for(int i=1;i<=N;i++)
{
fa[i]=i;
}
for(int i=1;i<=N;i++)
{
scanf("%d:",&K);
for(int j=1;j<=K;j++)
{
scanf("%d",&temp);
act[temp].a[++act[temp].num]=i;
}
}
for(int i=1;i<maxv;i++)
{
for(int j=2;j<=act[i].num;j++)
{
uni(act[i].a[1],act[i].a[j]);
}
}
for(int i=1;i<=N;i++)
{
root[findfa(i)]++;
}
int co=0;
for(int i=1;i<=N;i++)
{
if(root[i]>0)
co++;
}
printf("%d\n",co);
sort(root+1,root+1+N);
for(int i=N;i>N-co;i--)
{
if(i==N)
printf("%d",root[i]);
else
printf(" %d",root[i]);
}

return 0;
}