hdu 5425 Rikka with Tree II（暴力）

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题目链接：hdu 5425 Rikka with Tree II

直接枚举就好了，当概率极小时贡献值可以忽略。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

const int maxn = 1e5 + 5;

int N, D[maxn];
vector<int> G[maxn];

void init () {
	for (int i = 1; i <= N; i++) G[i].clear();

	int t;
	for (int i = 2; i <= N; i++) {
		scanf("%d", &t);
		G[t].push_back(i);
	}

	queue<int> Q;
	D[1] = 0;
	Q.push(1);

	while (!Q.empty()) {
		int u = Q.front();
		Q.pop();

		for (int i = 0; i < G[u].size(); i++) {
			int v = G[u][i];
			D[v] = D[u] + 1;
			Q.push(v);
		}
	}
	sort(D + 1, D + 1 + N);
}

int main () {
	while (scanf("%d", &N) == 1) {
		init ();

		double ans = 0;
		for (int i = 0; i < N && i < 100; i++) {
			int u = N - i;
			double w = 0;
			for (int j = N; j > u; j--)
				w += 1.0 * (D[u] + 1) * (D[j] + 1) / (D[u] + D[j] + 2);
			double t = pow(2, N-u+1) - (N+1) / pow(2, u-1);
			ans += w / t;
		}
		printf("%.6lf\n", ans);
	}
	return 0;
}

hdu 5425 Rikka with Tree II（暴力）

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/48143325