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,注意给出边是无向的就好了.

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;
}