Reverse algorithm set to --- Blue Bridge Cup

Problem Description
There are a binary 2n-1 nodes, it has exactly n leaf nodes, wrote an integer on each node. If the number of these on all leaf nodes of the tree from left to right written down, they will have a sequence of a [1] ... a [n ]. Now get this minimal sequence in reverse order of the number, but the only action is to choose a non-leaf node of the tree, it's about two subtrees exchange. He can do any number of times this operation.
In seeking the optimal solution, in reverse order of the sequence number of the minimum number required to complete in the shortest time.
The input format
of the first line of an integer n.
Each line below, a number x.
If x = 0, indicates that this node non-leaf nodes, recursively read down into its left child and right child's information, if x ≠ 0, indicates that this node is a leaf node, the right value of x.
Output format
output an integer that represents the minimum number of reverse right.
Sample input
. 3
0
0
. 3
1
2
Sample Output
1

#include<bits/stdc++.h>
using namespace std;
#define ForD(i,n) for(int i=n;i;i--)
#define F (100000007)
#define MAXN (2*200000+10)

long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}

int n,root=0;

struct node
{
    int fa,ch[2],size,c;
    node():size(0),c(0){ch[0]=ch[1]=fa=0;}
}a[MAXN];

void update(int x){a[x].size=a[a[x].ch[0]].size+a[a[x].ch[1]].size+(a[x].c>0);}
int tail=0;
void pushdown(int x){a[a[x].ch[0]].fa=a[a[x].ch[1]].fa=x;}

void build(int &x)
{
    if (!x) x=++tail;
    scanf("%d",&a[x].c);
    if (a[x].c==0)
    {
        build(a[x].ch[0]);
        build(a[x].ch[1]);
        update(x);pushdown(x);
    }else a[x].size=1;
}

void rotate(int x)
{
    int y=a[x].fa,z=a[y].fa;
    bool p=a[y].ch[0]==x;
    if (z)
    {
        if (a[z].ch[0]==y) a[z].ch[0]=x;
        else a[z].ch[1]=x;
    }
    a[x].fa=z,a[y].fa=x;
     if (a[x].ch[p]) a[a[x].ch[p]].fa=y;
    a[y].ch[p^1]=a[x].ch[p];
    a[x].ch[p]=y;
    update(y);
}

void splay(int x)
{
    while (a[x].fa)
    {
        int y=a[x].fa,z=a[y].fa;
        if (z)
            if ((a[y].ch[0]==x)^(a[z].ch[0]==y)) rotate(x);
            else rotate(y);
        rotate(x);
    }
    update(x);
}

void ins(long long &tot,int x,int y)
{
    a[x].size++;
    if (a[y].c<=a[x].c)
    {
        if (a[x].ch[0]) ins(tot,a[x].ch[0],y);
        else a[y].fa=x,splay(a[x].ch[0]=y);
    }
    else
    {
        tot+=a[a[x].ch[0]].size+(a[x].c>0);
        if (a[x].ch[1]) ins(tot,a[x].ch[1],y);
        else a[y].fa=x,splay(a[x].ch[1]=y);
    }
}

int q[MAXN],size;
void clac(int x,int y)
{
    if (a[y].ch[0]) clac(x,a[y].ch[0]);
    if (a[y].c) q[++size]=y;
    if (a[y].ch[1]) clac(x,a[y].ch[1]);
}

long long merge(bool &lor,int z)
{
    int x=a[z].ch[0],y=a[z].ch[1];
    if (a[x].size<a[y].size) swap(x,y);
 
    a[x].fa=0;a[y].fa=0;q[1]=y;
    size=0;clac(x,y);
   long long tot=0;
    ForD(i,size)
    {
        int now=q[i];
        a[now].ch[0]=a[now].ch[1]=a[now].fa=0;a[now].size=1;
        ins(tot,x,now);
        x=now;
    }
    a[x].fa=z;
    a[z].ch[0]=0,a[z].ch[1]=x;
    return tot;
}

long long qur(int &x)
{
    if (a[x].c) return 0;
    else
    {
        long long lson=a[a[x].ch[0]].size,rson=a[a[x].ch[1]].size,ls=qur(a[x].ch[0]),rs=qur(a[x].ch[1]);
        bool lor=0;
        long long ms=merge(lor,x);
        return ls+rs+min(lson*rson-ms,ms);
    }
}

int main()
{
    cin>>n;
    build(root);
    cout<<qur(root)<<endl;
    return 0;
}