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  • Valid BFS?

    D. Valid BFS?
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    The BFS algorithm is defined as follows.

    1. Consider an undirected graph with vertices numbered from 11 to nn. Initialize qq as a new queue containing only vertex 11, mark the vertex 11 as used.
    2. Extract a vertex vv from the head of the queue qq.
    3. Print the index of vertex vv.
    4. Iterate in arbitrary order through all such vertices uu that uu is a neighbor of vv and is not marked yet as used. Mark the vertex uu as used and insert it into the tail of the queue qq.
    5. If the queue is not empty, continue from step 2.
    6. Otherwise finish.

    Since the order of choosing neighbors of each vertex can vary, it turns out that there may be multiple sequences which BFS can print.

    In this problem you need to check whether a given sequence corresponds to some valid BFS traversal of the given tree starting from vertex 11. The tree is an undirected graph, such that there is exactly one simple path between any two vertices.

    Input

    The first line contains a single integer nn (1n21051≤n≤2⋅105) which denotes the number of nodes in the tree.

    The following n1n−1 lines describe the edges of the tree. Each of them contains two integers xx and yy (1x,yn1≤x,y≤n) — the endpoints of the corresponding edge of the tree. It is guaranteed that the given graph is a tree.

    The last line contains nn distinct integers a1,a2,,ana1,a2,…,an (1ain1≤ai≤n) — the sequence to check.

    Output

    Print "Yes" (quotes for clarity) if the sequence corresponds to some valid BFS traversal of the given tree and "No" (quotes for clarity) otherwise.

    You can print each letter in any case (upper or lower).

    Examples
    input
    Copy
    4
    1 2
    1 3
    2 4
    1 2 3 4
    output
    Copy
    Yes
    input
    Copy
    4
    1 2
    1 3
    2 4
    1 2 4 3
    output
    Copy
    No
    Note

    Both sample tests have the same tree in them.

    In this tree, there are two valid BFS orderings:

    • 1,2,3,41,2,3,4,
    • 1,3,2,41,3,2,4.

    The ordering 1,2,4,31,2,4,3 doesn't correspond to any valid BFS order.

    #include <bits/stdc++.h>
    
    using namespace std;
    typedef long long ll;
    const int maxn = 3e5+108;
    const ll mod = 1e9 + 7;
    int n;
    vector<int>adj[maxn],res;
    int seq[maxn],id[maxn];
    int vis[maxn];
    
    inline bool cmp(int s,int t){
        return id[s]<id[t];
    }
    
    int main() {
    #ifndef ONLINE_JUDGE
        freopen("1.txt", "r", stdin);
    #endif
        scanf("%d",&n);
        for(register int i=1,u,v;i<n;++i){
            scanf("%d%d",&u,&v);
            adj[u].emplace_back(v);
            adj[v].emplace_back(u);
        }
        for(register int i=1;i<=n;++i){
            scanf("%d",&seq[i]);
            id[seq[i]]=i;
        }
        for(register int i=1;i<=n;++i){
            sort(adj[i].begin(),adj[i].end(),cmp);
        }
        queue<int>q;
    //    q.push(1);
    //    vis[1]=1;
    //    while(!q.empty()){
    //        queue<int>pre;
    //        while(!q.empty()){
    //            int cur=q.front();
    //            q.pop();
    //            res.emplace_back(cur);
    //            for(register int i=0;i<adj[cur].size();++i){
    //                if(!vis[adj[cur][i]]){
    //                    pre.push(adj[cur][i]);
    //                    vis[adj[cur][i]]=1;
    //                }
    //            }
    //        }
    //        q=pre;
    //    }
        q.push(1);
        vis[1]=1;
        while(!q.empty()){
            int cur=q.front();
            q.pop();
            res.emplace_back(cur);
            for(register int i=0;i<adj[cur].size();++i){
                if(!vis[adj[cur][i]]){
                    q.push(adj[cur][i]);
                    vis[adj[cur][i]]=1;
                }
            }
        }
        for(register int i=1;i<=n;++i){
            if(res[i-1]!=seq[i]){
                puts("No");
                return 0;
            }
        }
        puts("Yes");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/czy-power/p/11466514.html
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