[贪心]ARC080 E

Description

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

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;}