CF697D Puzzles

题意：给一棵有根树，dfs这棵树时，走向每个子节点的概率相等，求这棵树每个点的dfs时间戳期望。

题解：考虑一个节点，其他兄弟节点对它有贡献当且仅当其他节点在它之前被遍历，概率是1/2的。

所以ans[x] = ans[fa] + 1 + sz[brother]  / 2;

#include<bits/stdc++.h>
using namespace std;
#define N 100000
#define LL long long

inline LL read() {
    LL x = 0, f = 1; char a = getchar();
    while(a < '0' || a > '9') { if(a == '-') f = -1; a = getchar(); }
    while(a >= '0' && a <= '9') x = x *10 + a - '0', a = getchar();
    return x * f;
}

int n, head[N + 5], sz[N + 5], cnt = 1;
double ans[N + 5];

struct edges{
    int to, next;
}e[2 * N + 5];

inline void insert(int x) {
    int u = x, v = read();
    e[++cnt] = (edges) {u, head[v]}; head[v] = cnt;
    e[++cnt] = (edges) {v, head[u]}; head[u] = cnt;
}

void dfs(int x, int fa){
    sz[x] = 1;
    for(int i = head[x]; i; i = e[i].next) {
        if(e[i].to == fa) continue;
        dfs(e[i].to, x)    ;
        sz[x] += sz[e[i].to];
    }
}

void Dfs(int x, int fa) {
    for(int i = head[x]; i; i = e[i].next) {
        if(e[i].to == fa) continue;        
        ans[e[i].to] = 1.0 * (sz[x] - 1 - sz[e[i].to]) / 2.0 + 1.0 + ans[x];
        Dfs(e[i].to, x);
    }
}

int main() {
    n = read();
    for(int i = 2; i <= n; i++) insert(i);
    dfs(1, 0);
    ans[1] = 1;
    Dfs(1, 0);
    for(int i = 1; i <= n; i++) printf("%.7lf
",ans[i]);
    return  0;
}