题意:n个人,每个人有个兴趣列表,有同样的至少一个兴趣的人在一个圈子里,求有多少个圈子,并把圈子按照人数从大到小输出人数。
思路:并查集维护即可。
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <vector>
#include <queue>
using namespace std;
const int MAX_N = 1010;
int par[MAX_N];
int n, k, x;
map<int, vector<int> > mp;
map<int, int> cnt;
vector<int> ans;
void init(int n) {
for (int i = 1; i <= n; i++) par[i] = i;
}
int find(int x) {
if (x == par[x]) return x;
return par[x] = find(par[x]);
}
void uni(int x, int y) {
x = find(x); y = find(y);
if (x == y) return ;
par[x] = y;
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
scanf("%d", &n);
init(n);
for (int i = 1; i <= n; i++) {
scanf("%d:", &k);
for (int j = 0; j < k; j++) {
scanf("%d", &x);
mp[x].push_back(i);
}
}
for (auto it = mp.begin(); it != mp.end(); it++) {
vector<int>& v = it->second;
if (v.size() == 0) continue;
int t = v[0];
for (int i = 1; i < v.size(); i++) {
uni(t, v[i]);
}
}
for (int i = 1; i <= n; i++) {
cnt[find(i)]++;
}
for (auto i = cnt.begin(); i != cnt.end(); i++) {
if (i->second > 0)
ans.push_back(i->second);
}
printf("%d\n", ans.size());
sort(ans.begin(), ans.end(), greater<int>());
for (int i = 0; i < ans.size(); i++) {
if (i > 0) printf(" ");
printf("%d", ans[i]);
}
return 0;
}