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;
}

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转载自blog.csdn.net/qq_39396954/article/details/79195476