[Codeforces 1037D] Valid BFS?

[Topic link]

Http://codeforces.com/problemset/problem/1037/D

Algorithm

First find the parent node of each point, the size of each subtrees tree

Then determine whether the BFS sequence is legal

Time complexity: O (N)

Code

#include <bits/stdc++.h>using namespacestd;Const intMAXN = 2e5 +Ten;structedge{intto, NXT;} E[MAXN<<1];intn, tot;inta[maxn],fa[maxn],head[maxn],size[maxn];template<typename t> InlinevoidChkmax (T &x,t y) {x =Max (x, y);} Template<typename t> InlinevoidChkmin (T &x,t y) {x =min (x, y);} Template<typename t> InlinevoidRead (T &x) {T F=1; x =0; Charc =GetChar ();  for(;!isdigit (c); c = GetChar ())if(c = ='-') F =-F;  for(; IsDigit (c); c = GetChar ()) x = (x <<3) + (x <<1) + C-'0'; X*=F;} InlinevoidAddedge (intUintv) {Tot++; E[tot]=(Edge) {V,head[u]}; Head[u]=tot;}intMain () {read (n); if(n = =1) {puts ("YES");return 0; }  for(inti =1; I < n; i++)        {                intu, v; Read (u);                Read (v);                Addedge (U,V);        Addedge (V,u); }         for(inti =1; I <= N; i++) read (A[i]); if(a[1] !=1) {puts ("NO");return 0; } Queue<int>Q; fa[1] = -1; Q.push (1);  while(!Q.empty ()) {                intU =Q.front ();                Q.pop ();  for(inti = Head[u]; I i =e[i].nxt) {                        intv =e[i].to; if(V! =Fa[u]) {Size[u]++; FA[V]=u;                            Q.push (v); }                }                }        intt =1, cnt =0;  for(inti =2; I <= N; i++)        {                if(Fa[a[i]] = =A[t]) {CNT++; Continue; }                if(CNT = =Size[a[t]]) {                         while(T < i && size[a[t +1]] ==0) t++; if(T = =i) {printf ("no\n"); return 0; } CNT=1; T++; Continue; } printf ("no\n"); return 0; }        if(cnt = = Size[a[t]]) printf ("yes\n"); Elseprintf"no\n"); return 0; }

[Codeforces 1037D] Valid BFS?