[贪心]ARC080 E

贪心]ARC080 E"/>

[贪心]ARC080 E

Description $Description$

给一个排列 pn $p_n$，每次可以选择排列 p $p$中的相邻的两个数，把这两个数放到q$q$的最前面。
要求最小化字典序

Solution $Solution$

这不就是贪心啊。。
跟超级钢琴一样的做法。
因为这两个数在原排列中的奇偶性不同，开两颗线段树（好吧，是不想打ST表）存一下最小值的位置就好了。。

#include <bits/stdc++.h>
using namespace std;const int N = 202020;inline char get(void) {static char buf[100000], *S = buf, *T = buf;if (S == T) {T = (S = buf) + fread(buf, 1, 100000, stdin);if (S == T) return EOF;}return *S++;
}
template<typename T>
inline void read(T &x) {static char c; x = 0; int sgn = 0;for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';if (sgn) x = -x;
}int a[N];
int n, l, r;
struct Seg {int c[N];int mn[N << 2];inline int MinPos(int a, int b) {return c[a] < c[b] ? a : b;}inline void Build(int o, int l, int r) {if (l == r) return (void)(mn[o] = l);int mid = (l + r) >> 1;Build(o << 1, l, mid);Build(o << 1 | 1, mid + 1, r);mn[o] = MinPos(mn[o << 1], mn[o << 1 | 1]);}inline void Init(int fl) {for (int i = 0; i <= n; i++) c[i] = N;for (int i = fl; i <= n; i += 2)c[i] = a[i];Build(1, 1, n);}inline int MinPos(int o, int l, int r, int L, int R) {if (l >= L && r <= R) return mn[o];int mid = (l + r) >> 1, res = 0;if (L <= mid) res = MinPos(res, MinPos(o << 1, l, mid, L, R));if (R > mid) res = MinPos(res, MinPos(o << 1 | 1, mid + 1, r, L, R));return res;}
};
Seg S[2];
struct cpp {int l, r, s, t, fl;cpp(void) {}cpp(int _l, int _r, int f) {l = _l; r = _r; fl = f;s = S[fl].MinPos(1, 1, n, l, r);t = S[fl ^ 1].MinPos(1, 1, n, s, r);}inline bool operator <(const cpp &x) const{return a[s] > a[x.s];}
};
cpp x;
priority_queue<cpp> Q;int main(void) {read(n);for (int i = 1; i <= n; i++) read(a[i]);S[1].Init(1); S[0].Init(2);Q.push(cpp(1, n, 1));while (!Q.empty()) {x = Q.top(); Q.pop();printf("%d %d ", a[x.s], a[x.t]);if (x.l < x.s) Q.push(cpp(x.l, x.s - 1, x.fl));if (x.t < x.r) Q.push(cpp(x.t + 1, x.r, x.fl));if (x.s + 1 < x.t) Q.push(cpp(x.s + 1, x.t - 1, x.fl ^ 1));}return 0;
}