## Pat(A) 1107. Social Clusters (30)

### 1107. Social Clusters (30)

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 hi ... 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

### 解题报告

ans[i]表示以i为根的一簇的人数，ans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 = {0};
int pre = {0}; // pre[i] means the last person who has a hobby i
int 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++;
ans[getF(i)] ++;
}
sort(ans.begin() + 1,ans.end());
}

int main(){
init();
cal();
printf("%d\n",ans);
if(ans){
printf("%d",ans[ans.size() - 1]);
for(int i = ans.size() - 2,k = 1; k < ans; i--,k++)
printf(" %d",ans[i]);
printf("\n");
}
system("pause");;
}

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