Title Description
Trees are a very common data structure.
We call a connected undirected graph with N points and N-1 edges a tree.
If a point is taken as a root and traversed from the root, the other points have a precursor, and the tree becomes a rooted tree.
For two trees T1T_1T1 and T2T_2T2, if all points of the tree T1T_1T1 can be re-labeled so that the tree T1T_1T1 and the tree T2T_2T2 are identical, then the two trees are isomorphic. That is to say, they have the same shape.
Now, give you M rooted trees. Please divide them into several equivalent classes according to their isomorphism.
Input format
The first line, an integer M.
Next, M lines, each containing several integers, represent a tree. The first integer N represents points. Next, N integers, in turn, represent the number of the father node of each point numbered 1 to N. The father node number of the root node is 0.
Output format
Output M lines, an integer per line, representing the smallest number of trees isomorphic to each tree.
Input and Output Samples
Input #1
4
4 0 1 1 2
4 2 0 2 3
4 0 1 1 1
4 0 1 2 3
Output #1
1
1
3
1
Note/hint
Trees numbered 1, 2, 4 are isomorphic. A tree numbered 3 is isomorphic only to itself.
In 100% of the data, 1 < N,M < 501 Leq N,M Leq 501 N and M < 50.
#include<stdio.h>
struct node{
int ch[2];
int size;
int val;
};
node nn[4096];
int num;
void add(char* s,int n,int now){
nn[now].size++;
if(n==0){
nn[now].val++;
return;
}
if(!nn[now].ch[s[0]-'0']){
num++;
nn[num].ch[0]=0;
nn[num].ch[1]=0;
nn[num].size=0;
nn[num].val=0;
nn[now].ch[s[0]-'0']=num;
}
add(s+1,n-1,nn[now].ch[s[0]-'0']);
}
int query(char* s,int n,int now,int i){
if(n==0){
if(nn[now].val==0) nn[now].val=i;
return nn[now].val;
}
if(!nn[now].ch[s[0]-'0']){
num++;
nn[num].ch[0]=0;
nn[num].ch[1]=0;
nn[num].size=0;
nn[num].val=0;
nn[now].ch[s[0]-'0']=num;
}
return query(s+1,n-1,nn[now].ch[s[0]-'0'],i);
}
void sol(char* s,int n){
if(n==0) return;
int lastwz=0;
int cnt=0;
for(int i=0;i<n;i++){
if(s[i]=='0') cnt++;
else cnt--;
if(cnt==0){
sol(s+lastwz+1,i-lastwz-1);
lastwz=i+1;
}
}
num=0;
nn[num].ch[0]=0;
nn[num].ch[1]=0;
nn[num].size=0;
nn[num].val=0;
cnt=0;
lastwz=0;
for(int i=0;i<n;i++){
if(s[i]=='0') cnt++;
else cnt--;
if(cnt==0){
add(s+lastwz,i-lastwz+1,0);
lastwz=i+1;
}
}
int now=0;
cnt=0;
while(1){
if(nn[now].size==0) break;
nn[now].size--;
if(nn[now].val){
nn[now].val--;
now=0;
continue;
}
if(nn[now].ch[0]&&nn[nn[now].ch[0]].size){
s[cnt]='0';
now=nn[now].ch[0];
}
else{
s[cnt]='1';
now=nn[now].ch[1];
}
cnt++;
}
}
char s[64][128];
int len[64];
int head[64],last[128],to[128],cnt=0;
int size[64];
void add(int u,int v){
cnt++;
last[cnt]=head[u];
head[u]=cnt;
to[cnt]=v;
}
void dfs(int u,int f){
size[u]=1;
for(int i=head[u];i;i=last[i]){
int v=to[i];
if(v==f) continue;
dfs(v,u);
size[u]+=size[v];
}
}
int findroot(int u,int f,int n){
for(int i=head[u];i;i=last[i]){
int v=to[i];
if(v==f) continue;
if(size[v]<=n) continue;
return findroot(v,u,n);
}
return u;
}
int tim;
void dfs2(int u,int f,int id){
s[id][tim]='0';
tim++;
for(int i=head[u];i;i=last[i]){
int v=to[i];
if(v==f) continue;
dfs2(v,u,id);
}
s[id][tim]='1';
tim++;
}
int main(){
int m;
scanf("%d",&m);
for(int i=1;i<=m;i++){
int n;
scanf("%d",&n);
for(int j=1;j<=n;j++) head[j]=0;
cnt=0;
for(int j=1;j<=n;j++){
int v;
scanf("%d",&v);
if(v){
add(j,v);
add(v,j);
}
}
dfs(1,0);
int u=findroot(1,0,n>>1);
int v=0;
for(int j=head[u];j;j=last[j]){
if((size[to[j]]<<1)==n){
v=to[j];
break;
}
}
tim=0;
dfs2(u,v,i);
if(v) dfs2(v,u,i);
sol(s[i],tim);
len[i]=tim;
}
num=0;
nn[num].ch[0]=0;
nn[num].ch[1]=0;
nn[num].size=0;
nn[num].val=0;
for(int i=1;i<=m;i++) printf("%d\n",query(s[i],len[i],0,i));
return 0;
}