Codeforces Round #428 (Div. 2) C. Journey (简单搜索)

• 题意:给你一颗树(边是无向的),从根节点向下走,统计走到每个子节点的概率,求所有叶子节点的深度乘上概率的和.

• 题解:每层子节点的概率等于上一层节点的概率乘\(1\)除以这层的子节点数,所以我们用\(dfs\)或者\(bfs\)都可以写,其实就是个搜索裸题,注意给的边是无向的就好了.

• 代码:

  1.dfs:

  int n;
  int a,b;
  vector<int> v[N];
  double ans;
  
  void dfs(int u,int fa,double p,int len){
      if((int)v[u].size()==1 && u!=1){
          ans+=p*len;len;
          return;
      }
      for(auto w:v[u]){
          double p1;
          if(w==fa) continue;
          if(u==1)  p1=p*(1.0/(double)(v[u].size()));
          else p1=p*(1.0/(double)(v[u].size()-1));
          dfs(w,u,p1,len+1);
      }
  }
  
  int main() {
      //ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      scanf("%d",&n);
      for(int i=1;i<n;++i){
          scanf("%d %d",&a,&b);
          v[a].pb(b);
          v[b].pb(a);
      }
  
      dfs(1,-1,1.0,0);
  
      printf("%.6f\n",ans);
  
      return 0;
  }
  

  2.bfs

  struct misaka{
      int node;
      int fa;
      double p;
      double len;
  }e;
  
  int n;
  int a,b;
  vector<int> v[N];
  double ans;
  
  void bfs(){
      queue<misaka> q;
      e.node=1;e.p=1.0;e.len=0.0;e.fa=-1;
      q.push(e);
  
      while(!q.empty()){
          auto tmp=q.front();
          q.pop();
  
          int node=tmp.node;
          double p=tmp.p;
          double len=tmp.len;
          int fa=tmp.fa;
  
          if((int)v[node].size()==1 && node!=1){
              ans+=p*len;
          }
  
          for(auto w:v[node]){
              if(w==fa) continue;
              e.node=w;
              if(node==1)
              e.p=p*((double)1.0/(double)v[node].size());
              else e.p=p*(1.0/(double)(v[node].size()-1));
              e.len=len+1.0;
              e.fa=node;
              q.push(e);
          }
      }
  }
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      cin>>n;
      e.len=1.0;
      e.p=1.0;
      for(int i=1;i<n;++i){
          cin>>a>>b;
          v[a].pb(b);
          v[b].pb(a);
      }
  
      bfs();
  
     printf("%.6f\n",ans); 
  
      return 0;
  }