ZOJ3963-Heap Partition

A sequence S = {s₁, s₂, ..., s_n} is called heapable if there exists a binary tree T with n nodes such that every node is labelled with exactly one element from the sequence S, and for every non-root node s_i and its parent s_j, s_j ≤ s_i and j < i hold. Each element in sequence S can be used to label a node in tree T only once.

Chiaki has a sequence a₁, a₂, ..., a_n, she would like to decompose it into a minimum number of heapable subsequences.

Note that a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contain an integer n (1 ≤ n ≤ 10⁵) — the length of the sequence.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n).

It is guaranteed that the sum of all n does not exceed 2 × 10⁶.

Output

For each test case, output an integer m denoting the minimum number of heapable subsequences in the first line. For the next m lines, first output an integer C_i, indicating the length of the subsequence. Then output C_i integers P_i1, P_i2, ..., P_iCi in increasing order on the same line, where P_ij means the index of the j-th element of the i-th subsequence in the original sequence.

Sample Input

4
4
1 2 3 4
4
2 4 3 1
4
1 1 1 1
5
3 2 1 4 1

Sample Output

1
4 1 2 3 4
2
3 1 2 3
1 4
1
4 1 2 3 4
3
2 1 4
1 2
2 3 5

Hint

题意 ： 构造尽可能少的一种结构，父节点的值小于等于子节点，而且子节点在序列中出现在父节点后面。而且每个父节点至多有两个子节点。

解题思路：线段树+离散化，判断每个数下面可以接几个儿子节点

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n, visit[100050], b[100050], sum[100050 * 4];
struct node
{
	int id, x;
}ans[100050];

void build(int k, int l, int r)
{
	sum[k] = 0;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(k << 1, l, mid);
	build(k << 1 | 1, mid + 1, r);
}

int query(int k, int l, int r, int ll, int rr)
{
	if (sum[k] == 0) return 0;
	if (l >= ll&&r <= rr)
	{
		if (l == r) return l;
		int mid = (l + r) >> 1, ma;
		if ((ma = query(k << 1 | 1, mid + 1, r, ll, rr))) return ma;
		else return query(k << 1, l, mid, ll, rr);
	}
	else
	{
		int mid = (l + r) >> 1, ma;
		if (mid < rr && (ma = query(k << 1 | 1, mid + 1, r, ll, rr))) return ma;
		else return query(k << 1, l, mid, ll, rr);
	}
}

void add(int k, int l, int r, int x, int val)
{
	if (l == r)
	{
		sum[k] += val;
		return;
	}
	int mid = (l + r) >> 1;
	if (x <= mid) add(k << 1, l, mid, x, val);
	else add(k << 1 | 1, mid + 1, r, x, val);
	sum[k] = max(sum[k << 1], sum[k << 1 | 1]);
}

bool cmp(node a, node b)
{
	if (a.x != b.x) return a.x < b.x;
	else return a.id < b.id;
}

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int xx, cnt = 1;
		scanf("%d%d", &n, &xx);
		for (int i = 1; i <= n; i++) visit[i] = 0;
		build(1, 1, n);
		ans[1].id = 1, ans[1].x = 1;
		visit[1]++, b[xx] = 1;
		add(1, 1, n, xx, 2);
		for (int i = 2; i <= n; i++)
		{
			scanf("%d", &xx);
			int k = query(1, 1, n, 1, xx);
			ans[i].id = i;
			k == 0 ? ans[i].x = ++cnt : ans[i].x = b[k];
			add(1, 1, n, xx, 2);
			if (k != 0) add(1, 1, n, k, -1);
			visit[ans[i].x]++, b[xx] = ans[i].x;
		}
		sort(ans + 1, ans + 1 + n, cmp);
		printf("%dn", cnt);
		printf("%d %d", visit[1], ans[1].id);
		for (int i = 2; i <= n; i++)
		{
			if (ans[i].x == ans[i - 1].x) printf(" %d", ans[i].id);
			else printf("n%d %d", visit[ans[i].x], ans[i].id);
		}
		printf("n");
	}
	return 0;
}

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