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HDU 5338 ZZX AND PERMUTATIONS 线段树

2024-09-23 15:13:59   215人阅读

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多校题解

胡搞。。

。

题意太难懂了。

。

ZZX and Permutations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 310    Accepted Submission(s): 83


Problem Description
ZZX likes permutations.

ZZX knows that a permutation can be decomposed into disjoint cycles(see https://en.wikipedia.org/wiki/Permutation#Cycle_notation). For example:
145632=(1)(35)(462)=(462)(1)(35)=(35)(1)(462)=(246)(1)(53)=(624)(1)(53)……
Note that there are many ways to rewrite it, but they are all equivalent.
A cycle with only one element is also written in the decomposition, like (1) in the example above.

Now, we remove all the parentheses in the decomposition. So the decomposition of 145632 can be 135462,462135,351462,246153,624153……

Now you are given the decomposition of a permutation after removing all the parentheses (itself is also a permutation). You should recover the original permutation. There are many ways to recover, so you should find the one with largest lexicographic order.
 

Input
First line contains an integer t, the number of test cases.
Then t testcases follow. In each testcase:
First line contains an integer n, the size of the permutation.
Second line contains n space-separated integers, the decomposition after removing parentheses.

n≤105. There are 10 testcases satisfying n≤105, 200 testcases satisfying n≤1000.
 

Output
Output n space-separated numbers in a line for each testcase.
Don‘t output space after the last number of a line.
 

Sample Input
2
6
1 4 5 6 3 2
2
1 2
 

Sample Output
4 6 2 5 1 3
2 1
 

Author
XJZX
 

Source
2015 Multi-University Training Contest 4
#include <iostream>
#include <fstream>
#include <string.h>
#include <string>
#include <time.h>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <stack>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
using namespace std;
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x < 0) {
		putchar('-');
		x = -x;
	}
	if (x > 9) pt(x / 10);
	putchar(x % 10 + '0');
}
typedef long long ll;
typedef pair<int, ll> pii;
const double eps = 1e-9;
const int N = 200000 + 10;
#define L(x) tree[x].l
#define R(x) tree[x].r
#define M(x) tree[x].ma
#define ls (id<<1)
#define rs (id<<1|1)
struct node {
	int l, r;
	int ma;
}tree[N << 2];
int a[N], p[N];
void Up(int id) {
	M(id) = max(M(ls), M(rs));
}
void build(int l, int r, int id) {
	L(id) = l; R(id) = r;
	if (l == r) { M(id) = a[l];return; }
	int mid = (l + r) >> 1;
	build(l, mid, ls); build(mid + 1, r, rs);
	Up(id);
}
void update(int pos, int id) {
	if (L(id) == R(id))
	{
		M(id) = -1;return;
	}
	int mid = (L(id) + R(id)) >> 1;
	if (pos <= mid)update(pos, ls);
	else update(pos, rs);
	Up(id);
}
int query(int l, int r, int id) {
	if (l == L(id) && R(id) == r)return M(id);
	int mid = (L(id) + R(id)) >> 1;
	if (r <= mid)return query(l, r, ls);
	else if (mid < l)return query(l, r, rs);
	else return max(query(l, mid, ls), query(mid + 1, r, rs));
}
int n;
int use[N], num[N];
pii b[N];
int ans[N];
void getcir(int l, int r) {
	if (l > r)return;
	for (int i = l; i <= r; i++) {
		if (use[a[i]])continue;
		int to = i + 1;
		if (to > r) to = l;
		ans[a[i]] = a[to];
		use[a[i]] = 1;
		num[a[to]] = 1;
		update(i, 1);
	}
}
int getmax(int l, int r) {
	if (l > r)return -1;
	return query(l, r, 1);
}
int hehe[N];
set<int>s;
int main() {
	int T;rd(T);
	while (T--) {
		s.clear();
		s.insert(0);
		rd(n);
		for (int i = 1; i <= n; i++) {
			rd(a[i]);
			p[a[i]] = i;
			use[i] = num[i] = false;
			b[i] = { a[i], i };
			ans[i] = 0;
		}
		build(1, n, 1);
		sort(b + 1, b + 1 + n);
		int top = 0;
		for (int i = 1; i <= n; i++) {
			if (use[i])continue;
			int idx = b[i].second;
			int t[3] = { -1, -1, -1 };
			if (idx < n && !num[a[idx+1]])t[0] = a[idx + 1];
			top = -(*s.upper_bound(-idx));
			t[1] = getmax(top + 1, idx - 1);
			if (num[i]==false)t[2] = i;
			if (t[0] > max(t[1], t[2]))
			{
				ans[i] = t[0]; use[i] = 1;
				num[t[0]] = 1; 
				update(idx + 1, 1);
			}
			else if (t[1] > max(t[0], t[2]))
			{
				getcir(p[t[1]], idx);
				s.insert(-idx);
			}
			else {
				getcir(idx, idx);
				s.insert(-idx);
			}
		}
		for (int i = 1; i <= n; i++)
		{
			pt(ans[i]);i == n ? putchar('\n') : putchar(' ');
		}
	}
	return 0;
}
/*
99
3
1 3 2
ans: 3 2 1
5
1 5 2 3 4
ans: 5 3 4 2 1
 
5
5 2 3 4 1
ans : 5 3 4 1 2
 
7
6 7 1 3 2 5 4
ans:7 5 3 4 2 6 1
 
1
8
1 3 6 4 8 7 2 5
 
1
5
3 2 4 5 1
 
*/


HDU 5338 ZZX AND PERMUTATIONS 线段树


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