http://codeforces.com/problemset/problem/839/C Journey

dfs跑树，每次记录路径个数及到达当前结点的概率，期望值等于到达各个叶子结点的步数*概率值的和，注意n=1时的特判。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <list>
#include <stack>
#include <cstring>
#include <queue>
#include <map>
#include <string>
using namespace std;
#define MAX 205
#define ll long long
#define INF 0x3f3f3f3f
#define mod 1000000007
vector<int> G[100005];
bool vis[100005];
double p, ans;
void dfs(int c, int v,double p) {
    vis[v] = true;
    bool f = false;
    int si = G[v].size() - 1;
    if (si) { 
        if(v != 1) p /= si; 
        else p /= (si + 1);
    }
    for (int i = 0; i <= si; ++i) {
        int t = G[v][i];
        if (!vis[t]) {
            f = true;
            dfs(c + 1, t,p);
        }
    }
    if (!f) {
        ans = ans + c * p;
    }
}
int main()
{
    freopen("a.txt", "r", stdin);
    freopen("b.txt", "w", stdout);
    int n, x, y;
    cin >> n;
    if (n == 1) {
        printf("0\n");
        return 0;
    }
    for (int i = 1; i < n; ++i) {
        scanf("%d%d", &x, &y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    dfs(0, 1,1);
    printf("%.15lf\n", ans);
    return 0;
}