L3-016 二叉搜索树的结构

输入样例：
5
2 4 1 3 0
8
2 is the root
1 and 4 are siblings
3 and 0 are on the same level
2 is the parent of 4
3 is the left child of 4
1 is the right child of 2
4 and 0 are on the same level
100 is the right child of 3

输出样例：
Yes
Yes
Yes
Yes
Yes
No
No
No

---

题解 : 写的代码极丑。
先离散化,建树的时候储存下来这个节点对应到数组上的下标，a[id[i]]=x，i是x的映射。上次有一个题目也是这样的查询,但是那个是通过每次把所有的点for循环直至a[i]=x，找到对应的数组下标,时间复杂度较高。

下面的代码是错误的，最后一个测试点MLE。

#include<cstdio>
#include<map>
using namespace std;
const int N=100+5,M=N*N*1000,INF=0x3f3f3f3f;
map<int,int>refl;
int root,k[N],a[M],id[N],tot,vis[M];
void insert_(int x,int h,int pos)
{
    if(!vis[pos])
    {
        vis[pos]=1;
        a[pos]=x;
        if(!refl[x])
            refl[x]=++tot;
        id[tot]=pos;
        k[tot]=h;
        return;
    }
    if(x<a[pos])
        insert_(x,h+1,pos<<1);
    else
        insert_(x,h+1,pos<<1|1);
    k[pos]=h;
}
int main()
{
    tot=0;
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        int x;scanf("%d",&x);
        insert_(x,1,1);
    }
    int m;scanf("%d",&m);
    while(m--)
    {
        int x,y,tmp;char s[15];
        scanf("%d%s",&x,s);
        tmp=x,x=refl[x];
        if(s[0]=='a')
        {
            scanf("%d",&y);
            scanf("%s",s),scanf("%s",s);
            y=refl[y];
            if(s[0]=='s')//siblings
            {
                if(!x||!y||x==y||id[x]==1||id[y]==1)
                    {puts("No");continue;}
                puts(a[id[x]>>1]==a[id[y]>>1]?"Yes":"No");
            }
            else//the same level
            {
                scanf("%s",s),scanf("%s",s),scanf("%s",s);
                if(!x||!y)
                    {puts("No");continue;}
                puts(k[x]==k[y]?"Yes":"No");
            }
        }
        else
        {
            scanf("%s",s),scanf("%s",s);
            if(s[0]=='r'&&s[1]=='o')//root
            {
                if(!x)
                    {puts("No");continue;}
                puts(tmp==a[1]?"Yes":"No");
                continue;
            }
            if(s[0]=='l')//left child
            {
                scanf("%s",s),scanf("%s%d",s,&y);
                if(!refl[y]||!x||id[x]==1)
                    {puts("No");continue;}
                puts((!(id[x]&1)&&a[id[x]>>1]==y)?"Yes":"No");
            }
            if(s[0]=='r')//right child
            {
                scanf("%s",s),scanf("%s%d",s,&y);
                if(!refl[y]||!x||id[x]==1)
                    {puts("No");continue;}
                puts(((id[x]&1)&&a[id[x]>>1]==y)?"Yes":"No");
            }
            if(s[0]=='p')//parent
            {
                scanf("%s%d",s,&y);
                y=refl[y];
                if(!y||!x||id[y]==1)
                    {puts("No");continue;}
                puts(tmp==a[id[y]>>1]?"Yes":"No");
            }
        }
    }
}