题目：http://community.topcoder.com/stat?c=problem_statement&pm=13117&rd=15950

看到树，又是与节点有关，通常是dp，dp[v][t] 表示在以v为root的subtree上有t个eagle时满足条件的概率。一个注意点是求二项系数数组C[]时，由于值太大。要用double，不能用long long, long long 会溢出。

#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <bitset>
#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <map>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>
using namespace std;

#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)
typedef pair<int, int> pii;
typedef long long llong;
typedef pair<llong, llong> pll;
#define mkp make_pair

/*************** Program Begin **********************/
const int MAX_N = 51, MAX_K = 101;
double dp[MAX_N][MAX_K];
double C[MAX_K][MAX_K];		// 二项系数，注意：用 long long 会溢出
class EagleInZoo {
public:
	vector < vector<int> > childs;
	double rec(int v, int t)
	{
		if (1 == t) {
			return 1.0;
		}
		double & res = dp[v][t];
		if (res > -0.5) {
			return res;
		}
		if (childs[v].size() == 0 && t > 1) {	// 该节点无 child 且t > 1。最后一仅仅一定无法停留
			res = 0;
			return res;
		}
		res = 0;

		for (int i = 0; i < childs[v].size(); i++) {
			int x = childs[v][i];
			int c = childs[v].size();
			for (int j = 0; j <= t - 2; j++) {
				res += (1.0 / c) * C[t-2][j] * pow(1.0 / c, j) * pow( 1.0 * (c-1) / c, t-2-j) * rec(x, j + 1);
			}
		}

		return res;
	}
	double calc(vector <int> parent, int K) {
		for (int i = 0; i < MAX_N; i++) {
			for (int j = 0; j < MAX_K; j++) {
				dp[i][j] = -1.0;
			}
		}
		childs.resize(parent.size() + 1);
		for (int i = 0; i < parent.size(); i++) {
			childs[ parent[i] ].push_back(i + 1);
		}
		// calculate C[]
		C[0][0] = 1;
		for (int i = 1; i <= K - 1; i++) {
			C[i][0] = C[i][i] = 1;
			for (int j = 1; j < i; j++) {
				C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
			}
		}
		return rec(0, K);
	}

};

/************** Program End ************************/