Tree Metro System

topic

topic

Ideas

The first two are read in https://www.acwing.com/solution/content/4142/.

1

We definitely want to find the smallest representation of the two trees and see if they are the same.

In this case, we expect that each point in its subtree selects child nodes from large to small according to the longest chain length for DFS, but what if it is the same? We are going to do the same thing on the chain. In theory, I think memory search should be able to reach $O\left(n^2\right)$, but I think I can't type this code even more.

2

The optimization of the previous method is to directly find the minimum representation of the child node, and then put the minimum representation of the child node in lexicographical order, and then continue to merge upwards, the minimum representation will come out, and finally judge it.

They say it is $O\left(n^2\right)$, but because of the sorting, I think it is not$O\left(n^2\right)$It should be slightly larger, but it won’t get stuck. (The same is true for the first one above. But if you use a similar radix sorting method, it is indeed$O\left(n^2\right)$. )

3

My thinking is very violent.

$pd(x,y)$ means to judge$x,y$ Are the two subtrees the same? How to judge? We putxxThe child nodes of$x$ andyyThe child node of$y$ is judged, and finally if it is found to be matched one by one, it returns$1$ , otherwise return$0$。

But how to prove the time complexity?

Obviously, for two points at the same height, at most $pd$ once, because for$pd(x,y)$ ,xxchild nodeaa of$x$$a$和$y$ 's child node$b$ will only$pd$ once, regardless of success or failure, the code will no longer judge$pd(x,y)$ , so$a,b$ will onlypd pdeach other$pd$ once.

Using this, we can get the time complexity: $O(n^2)$

Violence works miracles! ! ! !

Code

#include<cstdio>
#include<cstring>
#define  N  6100
using  namespace  std;
struct  node
{
    
    
    int  y,next;
}a[N],b[N];int  len,last[N],tot/*点的个数*/;
int  n,fa[N],cnt[N];
void  insa(int  x,int  y)
{
    
    
    len++;
    a[len].y=y;a[len].next=last[x];last[x]=len;
}
void  insb(int  x,int  y)
{
    
    
    len++;
    b[len].y=y;b[len].next=last[x];last[x]=len;
}
char  st[N];
int  aa[N],bb[N];
int  sta[N],top;//sta栈 
int  v[N],ts;
bool  pd(int  x,int  y)
{
    
    
    if(cnt[x]!=cnt[y])return  0;
    ts++;int  now=ts;
    for(int  k=last[x];k;k=a[k].next)
    {
    
    
        int  xx=a[k].y,t=0;
        for(int  kk=last[y];kk;kk=b[kk].next)
        {
    
    
            int  yy=b[kk].y;
            if(v[yy]!=now)
            {
    
    
                if(pd(xx,yy)==1)
                {
    
    
                    t=1;
                    v[yy]=now;break;
                }
            }
        }
        if(t==0)
        {
    
    
            return  0;
        }
    }
    return  1;
}
int  main()
{
    
    
    int  T;scanf("%d",&T);
    while(T--)
    {
    
    
        tot=0;len=0;top=0;memset(last,0,sizeof(last));
        memset(a,0,sizeof(a));memset(b,0,sizeof(b));
        memset(cnt,0,sizeof(cnt));ts=0;memset(v,0,sizeof(v));
        //初始化
        scanf("%s",st+1);n=strlen(st+1); 
        for(int  i=1;i<=n;i++)aa[i]=st[i]-'0';
        scanf("%s",st+1);
        for(int  i=1;i<=n;i++)bb[i]=st[i]-'0';
        sta[top]=++tot;
        for(int  i=1;i<=n;i++)
        {
    
    
            if(aa[i]==0)
            {
    
    
                int  x=sta[top];
                sta[++top]=++tot;cnt[tot]=1;
                fa[tot]=x;insa(x,tot);//收一个儿子 
            }
            else  cnt[fa[sta[top]]]+=cnt[sta[top]],top--;
        }
        sta[top]=++tot;
        for(int  i=1;i<=n;i++)
        {
    
    
            if(bb[i]==0)
            {
    
    
                int  x=sta[top];
                sta[++top]=++tot;cnt[tot]=1;
                fa[tot]=x;insb(x,tot);//收一个儿子 
            }
            else  cnt[fa[sta[top]]]+=cnt[sta[top]],top--;
        }
        if(pd(1,tot/2+1))printf("same\n");
        else  printf("different\n");
    }
    return  0;
}