CodeForces 280C :Game on Tree 数学期望

传送门

题意

分析

每个点有两种情况，删除或者不删除，也就是对应操作0次或者操作1次

所以，我们考虑对每个点单独计算贡献，因为每个点操作次数最多为1，所以期望应该是每个点被选中的概率和，考虑一条从根结点到 u u u的链，什么情况下 u u u必选呢，当 1 − ( u − 1 ) 1 - (u - 1) 1−(u−1)这些点都没有被选中的情况下， u u u节点必选，所以概率为 1 / d e p 1 / dep 1/dep

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10,M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int h[N],e[M],ne[M],idx;
int n,k;
double ans;

void add(int x,int y){
    ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}

void dfs(int u,int fa,int dep){
    ans = ans + 1 / (1.0 * dep);
    for(int i = h[u];~i;i = ne[i]){
        int j = e[i];
        if(j == fa) continue;
        dfs(j,u,dep + 1);
    }
}

int main() {
    read(n);
    for(int i = 1;i <= n;i++) h[i] = -1;
    for(int i = 1;i < n;i++){
        int x,y;
        read(x),read(y);
        add(x,y),add(y,x);
    }
    printf("%.12lf\n",ans);
    return 0;

}