[Codeforces 1037D] Valid BFS？

[题目链接]

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

[算法]

         首先求出每个点的父节点 ， 每棵子树的大小

         然后判断BFS序是否合法即可

         时间复杂度 ： O(N)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;

struct edge
{
        int to , nxt;
} e[MAXN << 1];

int n , tot;
int a[MAXN],fa[MAXN],head[MAXN],size[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u,int v)
{
        tot++;
        e[tot] = (edge){v,head[u]};
        head[u] = tot;
}

int main()
{
        
        read(n);
        if (n == 1) { puts("YES"); return 0; }
        for (int i = 1; i < n; i++)
        {
                int u , v;
                read(u); read(v);
                addedge(u,v);
                addedge(v,u);
        }
        for (int i = 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())
        {
                int u = q.front();
                q.pop();
                for (int i = head[u]; i; i = e[i].nxt)
                {
                        int v = e[i].to;
                        if (v != fa[u])
                        {
                                size[u]++;
                                fa[v]    = u;
                                q.push(v);    
                        }
                }        
        }
        int t = 1 , cnt = 0;
        for (int i = 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");
        else printf("NO\n");
        
        return 0;
    
}