AtCoder Regular Contest 080 E

http://arc080.contest.atcoder.jp/tasks/arc080_c
其实细想其本质是一道拓扑序的题，选了当前点，才能选分裂出的三段。

#include<bits/stdc++.h>#define pb push_back #define mp make_pairusing namespace std;const int maxn=2e5+7;typedef long long ll;struct My{    int l,r;    pair<int,int> p;    My(int L,int R,pair<int,int>P)        :l(L),r(R),p(P) {}};bool operator<(const My &a,const My &b){    if(a.p>b.p) return 1;    return 0;}priority_queue<My>q;vector<pair<int,int> >ans;vector<int>b1,b2;int n;int h1,h2;int a[maxn];int pos[maxn];//int f[maxn];struct node{    int l,r;    int min;    void update(int v)    {        min+=v;    }}tree[maxn*4],tree2[maxn*4];void push_up(int o){    tree[o].min=min(tree[o<<1].min,tree[o<<1|1].min);}void build(int l,int r,int o){    tree[o].l=l;tree[o].r=r;    if(l==r)    {        tree[o].min=b1[l];    }    else    {        int mid=(l+r)/2;        build(l,mid,o<<1);        build(mid+1,r,o<<1|1);        push_up(o);    }}void update(int p,int x,int o){}int qmin;void query(int ql,int qr,int o){    int l=tree[o].l,r=tree[o].r;    if(ql<=l&&r<=qr)    {        qmin=min(qmin,tree[o].min);    }    else    {        int mid=(l+r)/2;        if(ql<=mid) query(ql,qr,o<<1);        if(qr>mid) query(ql,qr,o<<1|1);    }}//////////////////////////////////////////////////////////////////////////////void push_up2(int o){    tree2[o].min=min(tree2[o<<1].min,tree2[o<<1|1].min);}void build2(int l,int r,int o){    tree2[o].l=l;tree2[o].r=r;    if(l==r)    {        tree2[o].min=b2[l];    }    else    {        int mid=(l+r)/2;        build2(l,mid,o<<1);        build2(mid+1,r,o<<1|1);        push_up2(o);    }}void update2(int p,int x,int o){}void query2(int ql,int qr,int o){    int l=tree2[o].l,r=tree2[o].r;    if(ql<=l&&r<=qr)    {        qmin=min(qmin,tree2[o].min);    }    else    {        int mid=(l+r)/2;        if(ql<=mid) query2(ql,qr,o<<1);        if(qr>mid) query2(ql,qr,o<<1|1);    }}int Q1(int l,int r){    l=(l+1)/2;    r=(r+1)/2;    qmin=1e9;    query(l,r,1);    return pos[qmin];}int Q2(int l,int r){    l=(l)/2;    r=(r)/2;    qmin=1e9;    query2(l,r,1);    return pos[qmin];}int num1,num2;pair<int,int> work(int l,int r){       if(l+1==r)    {        return mp(a[l],a[r]);    }    int i,j,k;    if(l%2==1)      num1=Q1(l,r);    else num1=Q2(l,r);    if(l%2==1) num2=Q2(num1+1,r);    else num2=Q1(num1+1,r);    //ans.pb(mp(a[num1],a[num2]));    /*    work(l,num1-1);    work(num2+1,r);    work(num1+1,num2-1);    */    return mp(a[num1],a[num2]);}void doit(){    q.push(My(1,n,work(1,n)));    while(!q.empty())    {        My now=q.top();        q.pop();        ans.pb(now.p);        int l,r,l1,r1;        l=now.l;        r=now.r;        l1=pos[now.p.first];        r1=pos[now.p.second];        if(l<l1-1) q.push(My(l,l1-1,work(l,l1-1)));        if(l1+1<r1-1) q.push(My(l1+1,r1-1,work(l1+1,r1-1)));        if(r1+1<r) q.push(My(r1+1,r,work(r1+1,r)));    }}int main(){    int i,j,k;    scanf("%d",&n);    b1.pb(0);    b2.pb(0);    for(i=1;i<=n;++i)    {        scanf("%d",&a[i]);        pos[a[i]]=i;        if(i&1)        {            b1.pb(a[i]);        }        else b2.pb(a[i]);    }     h1=b1.size()-1;    h2=b2.size()-1;    build(1,h1,1);    build2(1,h2,1);    doit();    int len=ans.size();    for(i=0;i<len-1;++i)    {        printf("%d %d ",ans[i].first,ans[i].second);    }    printf("%d %d\n",ans[len-1].first,ans[len-1].second);    return 0;}

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另解：斌大爷的分块