1106 Lowest Price in Supply Chain （25 分）

A supply chain is a network of retailers（零售商）, distributors（經銷商）, and suppliers（供應商）-- everyone involved in moving a product from supplier to customer.

Starting from one root supplier, everyone on the chain buys products from one's supplier in a price P and sell or distribute them in a price that is r% higher than P. Only the retailers will face the customers. It is assumed that each member in the supply chain has exactly one supplier except the root supplier, and there is no supply cycle.

Now given a supply chain, you are supposed to tell the lowest price a customer can expect from some retailers.

Input Specification:

Each input file contains one test case. For each case, The first line contains three positive numbers: N (≤10​5​​), the total number of the members in the supply chain (and hence their ID's are numbered from 0 to N−1, and the root supplier's ID is 0); P, the price given by the root supplier; and r, the percentage rate of price increment for each distributor or retailer. Then N lines follow, each describes a distributor or retailer in the following format:

K​i​​ ID[1] ID[2] ... ID[K​i​​]

where in the i-th line, K​i​​ is the total number of distributors or retailers who receive products from supplier i, and is then followed by the ID's of these distributors or retailers. K​j​​ being 0 means that the j-th member is a retailer. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the lowest price we can expect from some retailers, accurate up to 4 decimal places, and the number of retailers that sell at the lowest price. There must be one space between the two numbers. It is guaranteed that the all the prices will not exceed 10​10​​.

Sample Input:

10 1.80 1.00
3 2 3 5
1 9
1 4
1 7
0
2 6 1
1 8
0
0
0

Sample Output:

1.8362 2

解析: 大意就是根據輸入能構造出一棵樹，然後求葉子結點(對應零售商)出售標品價值的最小值。用dfs即可輕鬆解決。

注意：(1)res用於存放最低價格，其初始值要設大些，避免出錯。 

           (2)double要用%lf輸入，但要用%f輸出。或者用cin來輸入double,用%f打印出double值

#include<iostream>
#include<cstdio>
#include<vector> 
#include<cmath>
using namespace std;

vector<vector<int>> vec;
int n;
double p,r;

double res = 999999999;//初始值要設大些 
int num = 0;

void dfs(int node,int h){
	if(vec[node].size() == 0){
		double temp = pow(1+r*0.01,h)*p;//r記得換成百分比形式 
		if(temp < res){
			res = temp;
			num = 1;
		}else if(temp == res)
			num++;	
		return;
	}
	for(int i=0;i<vec[node].size();i++)
		dfs(vec[node][i],h+1);
}

int main(){
	scanf("%d %lf %lf",&n,&p,&r);
	vec.resize(n);
	for(int i=0;i<n;i++){
		int a,b;
		cin>>a;
		for(int j=0;j<a;j++){
			cin>>b;
			vec[i].push_back(b);
		}	
	}
	dfs(0,0);
	printf("%.4f %d",res,num);
	
	return 0;
}