链接

多校题解

胡搞。。。

题意太难懂了。。

ZZX and Permutations

2
6
1 4 5 6 3 2
2
1 2

4 6 2 5 1 3
2 1

#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

原文：http://blog.csdn.net/qq574857122/article/details/47160131