Atcoder arc080E Young Maids(线段树+优先队列)

给出一个n排列，每次可以选择相邻的两个数字放在新的排列首部，问最后形成的新的排列字典序最小是?

考虑新排列的第一个数字，则应是下标为奇数的最小数，下标不妨设为i。第二个数字应该下标大于i且为偶数的最小数，不妨设为j。

那么这样就将[1,n]新分割成了三个区间[1,i-1]，[i+1,j-1]，[j+1,n]。

用线段树实现查询操作。用优先队列维护区间的最优值。

时间复杂度O(nlogn).

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FDR(i,a,n) for(int i=a; i>=n; --i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
inline int Scan() {
    int x=,f=; char ch=getchar();
    while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}
    return x*f;
}
inline void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin....
 
struct qnode{
   int l, r, vl, vr;
   qnode(int _l=, int _r=, int _vl=, int _vr=):l(_l),r(_r),vl(_vl),vr(_vr){}
   bool operator<(const qnode &m)const{return vl>m.vl;}
};
priority_queue<qnode>que;
struct Seg{int odd, even;}seg[N<<];
int a[N], id[N];
VI ans;
 
void push_up(int p){
    seg[p].even=min(seg[p<<].even, seg[p<<|].even);
    seg[p].odd=min(seg[p<<].odd, seg[p<<|].odd);
}
void init(int p, int l, int r){
    if (l<r) {
        int mid=(l+r)>>;
        init(lch); init(rch); push_up(p);
    }
    else {
        if (l&) seg[p].even=a[l], seg[p].odd=INF;
        else seg[p].odd=a[l], seg[p].even=INF;
    }
}
int query(int p, int l, int r, int L, int R, int flag){
    if (L>r||R<l) return INF;
    if (L<=l&&R>=r) return flag?seg[p].even:seg[p].odd;
    else {
        int mid=(l+r)>>;
        return min(query(lch,L,R,flag),query(rch,L,R,flag));
    }
}
int main ()
{
    int n, x, y, l, r, ll;
    qnode tmp;
    scanf("%d",&n);
    FOR(i,,n) scanf("%d",a+i), id[a[i]]=i;
    init(,,n);
    x=query(,,n,,n,); l=id[x]; y=query(,,n,l+,n,);
    que.push(qnode(,n,x,y));
    FOR(i,,n/) {
        tmp=que.top(); que.pop();
        ans.pb(tmp.vl); ans.pb(tmp.vr);
        l=id[tmp.vl]; r=id[tmp.vr];
        if (tmp.l<l-) {
            x=query(,,n,tmp.l,l-,tmp.l&); ll=id[x]; y=query(,,n,ll+,l-,(l-)&);
            que.push(qnode(tmp.l,l-,x,y));
        }
        if (l+<r-) {
            x=query(,,n,l+,r-,(l+)&); ll=id[x]; y=query(,,n,ll+,r-,(r-)&);
            que.push(qnode(l+,r-,x,y));
        }
        if (r+<tmp.r) {
            x=query(,,n,r+,tmp.r,(r+)&); ll=id[x]; y=query(,,n,ll+,tmp.r,tmp.r&);
            que.push(qnode(r+,tmp.r,x,y));
        }
    }
    for (int i=; i<ans.size(); ++i) printf(i==?"%d":" %d",ans[i]);
    putchar('\n');
    return ;
}