链接
多校题解
胡搞。。。
题意太难懂了。。
ZZX and PermutationsTime Limit: 6000/3000 MS (Java/Others)Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 310Accepted 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, the number of test cases.Thentestcases follow. In each testcase:First line contains an integer, the size of the permutation.Second line containsspace-separated integers, the decomposition after removing parentheses.. There are 10 testcases satisfying, 200 testcases satisfying.
Output
Outputspace-separated numbers in a line for each testcase.Don’t output space after the last number of a line.
Sample Input
261 4 5 6 3 221 2
Sample Output
4 6 2 5 1 32 1
Author
XJZX
Source
#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;}/*9931 3 2ans: 3 2 151 5 2 3 4ans: 5 3 4 2 1 55 2 3 4 1ans : 5 3 4 1 2 76 7 1 3 2 5 4ans:7 5 3 4 2 6 1 181 3 6 4 8 7 2 5 153 2 4 5 1 */
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，带着我的相机和电脑，远离繁华，走向空旷。
