codeforces#428（div2）C Journey（树上的dfs水题）

There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.

Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.

Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link https://en.wikipedia.org/wiki/Expected_value.

Input
The first line contains a single integer n (1 ≤ n ≤ 100000) — number of cities.

Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the cities connected by the i-th road.

It is guaranteed that one can reach any city from any other by the roads.

Output
Print a number — the expected length of their journey. The journey starts in the city 1.

Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .

Examples
input
4
1 2
1 3
2 4
output
1.500000000000000
input
5
1 2
1 3
3 4
2 5
output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.

In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.

题意：求一棵树上根节点到每个叶子的距离的期望（PS：一开始题目读错了，以为是求到各叶子距离的平均值，其实每个分叉点都有概率这一说的，如果有两个儿子那么到每个儿子的概率都是0.5，然后一路算下去 0.0，最后发现挺多人都错在这一组数据，就怀疑题目读错了，果然。。。在最后几分钟发现了。。。然后没时间改了，苦恼 ..0.0...）

思路：直接dfs出到叶子的距离和概率就好了

代码：

#include <cstdio>#include <cmath>#include <iostream>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <map>#include <numeric>#include <set>#include <string>#include <cctype>#include <sstream>#define INF 0x3f3f3f3f#define lson l, m, rt<<1#define rson m+1, r, rt<<1|1using namespace std;typedef long long LL;typedef pair<LL, LL> P;const int maxn = 1e5 + 5;const int mod = 1e8 + 7;int n,fa[maxn];vector<int>G[maxn];double cnt=0,sum=0;void dfs(int x,int d,double ss) {    if (G[x].size()==1&&x!=1) {        sum += (d-1)*ss;        return;    }    for (int i=0; i<G[x].size(); i++) {        int son=G[x][i];        if (son==fa[x]) continue;        fa[son]=x;        double sss ;        if(x == 1)          sss =   ss* 1.0/(G[x].size());        else sss = ss* 1.0/(G[x].size()-1);        dfs(son,d+1,sss);    }    return;}int main() {    //freopen ("in.txt", "r", stdin);    while (~scanf ("%d",&n)) {        if (n==1) {            printf ("0.000000000000000\n");            continue;        }        for (int i=0; i<=n; i++) G[i].clear();        cnt=0;        sum=0;        memset(fa,0,sizeof(fa));        int u,v;        n--;        while (n--) {            scanf ("%d%d",&u,&v);            G[u].push_back(v);            G[v].push_back(u);        }        dfs(1,1,1.0);        printf ("%.15lf\n",sum);    }    return 0;}