Codeforces Round #428C

Journey

题意：给一颗树，边权都为1，从1出发，求走到叶子节点的边权和的期望（每次往孩子遍历的等概率的）

思路：从1出发dfs，每个点到下一个点的概率是当前节点的概率乘孩子节点的个数，也就是当前点的边数-1，到叶子节点后计算概率乘边权和然后相加就是了，1节点需要特殊处理一下

AC代码：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("\n")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

vector<int> vex[N];
double dfs(int u, int fa, int l, double p){
    if(vex[u].size()==1){//cout<<p*l<<endl;
        return p*l;
    }
    double ret=0.0;
    for(auto v : vex[u]){ //cout<<u<<" "<<vex[u].size()<<endl;
        if(v!=fa) ret+=dfs(v,u,l+1,p/(vex[u].size()-1
                                      ));
    }
    return ret;
}
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n; cin>>n;
    for(int i=1; i<n; ++i){
        int u,v; cin>>u>>v;
        vex[u].pb(v), vex[v].pb(u);
    }
    vex[1].pb(0);
    double ans=dfs(1,0,0,1.0);
    cout<<step(6)ans<<endl;
    return 0;
}