Codeforces 280C Game on Tree

题目链接

http://codeforces.com/problemset/problem/280/C

分析

设随机变量 $X$ 表示操作次数，$X_i$ 表示节点 $i$ 是否被直接操作（“是”为 $1$，“否”为 $0$）

显然 $X={\sum }_{1\le i\le n}{X}_i$，则 $E(X)={\sum }_{1\le i\le n}E(X_i)$

而一个节点被直接操作的概率等价于从仅包含该点及其祖先的集合中选出该点的概率，即 $\frac{1}{depth[i]}$

故最终答案为 ${\sum }_{1\le i\le n}\frac{1}{depth[i]}$

AC代码

#include 

inline int read() {
	int num = 0;
	char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	while (c >= '0' && c <= '9')
		num = num * 10 + c - '0', c = getchar();
	return num;
}

const int maxn = 1e5 + 5;

int head[maxn], eid;

struct Edge {
	int v, next;
} edge[2 * maxn];

inline void insert(int u, int v) {
	edge[++eid].v = v;
	edge[eid].next = head[u];
	head[u] = eid;
}

int depth[maxn];

void dfs(int u, int fa) {
	for (int p = head[u]; p; p = edge[p].next) {
		int v = edge[p].v;
		if (v == fa) continue;
		depth[v] = depth[u] + 1;
		dfs(v, u);
	}
}

int main() {
	int n = read();
	for (int i = 1; i <= n - 1; ++i) {
		int u = read(), v = read();
		insert(u, v), insert(v, u);
	}
	depth[1] = 1;
	dfs(1, 0);
	double ans = 0;
	for (int i = 1; i <= n; ++i) ans += 1.0 / depth[i];
	printf("%f", ans);
	return 0;
}