题目链接

uojUNR3B

题解

如果不输出方案，是有一个经典的三分做法的

但是要输出方案也是可以贪心的

设\(d[i]\)为\(i\)节点到最深的儿子的距离

贪心选择\(d[i]\)大的即可

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<vector>

#include<queue>

#include<cmath>

#include<map>

#define LL long long int

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define cls(s,v) memset(s,v,sizeof(s))

#define mp(a,b) make_pair<int,int>(a,b)

#define cp pair<int,int>

using namespace std;

const int maxn = 100005,maxm = 100005,INF = 0x3f3f3f3f;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}

	return flag ? out : -out;

}

priority_queue<cp> q;

vector<int> out[maxn];

int ls[maxn],rb[maxn],d[maxn];

int n,m,ans;

void dfs(int u){for (int k = ls[u]; k; k = rb[k]) dfs(k),d[u] = max(d[u],d[k] + 1);}

void work(){

	q.push(mp(d[1],1));

	int cnt = 0;

	while (cnt < n){

		ans++;

		for (int i = 1; i <= m; i++){

			if (q.empty()) break;

			out[ans].push_back(q.top().second); q.pop();

			cnt++;

		}

		for (unsigned int j = 0; j < out[ans].size(); j++){

			int u = out[ans][j];

			for (int k = ls[u]; k; k = rb[k])

				q.push(mp(d[k],k));

		}

	}

	printf("%d\n",ans);

	for (int i = 1; i <= ans; i++,puts("")){

		printf("%d ",out[i].size());

		for (unsigned int j = 0; j < out[i].size(); j++)

			printf("%d ",out[i][j]);

	}

}

int main(){

	n = read(); m = read(); int f;

	for (int i = 2; i <= n; i++){

		f = read(); rb[i] = ls[f]; ls[f] = i;

	}

	dfs(1);

	work();

	return 0;

}