1106 Lowest Price in Supply Chain (25分) C++

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1106 Lowest Price in Supply Chain

题目描述
解题思路
代码

题目描述

A supply chain is a network of retailers（零售商）, distributors（经销商）, and suppliers（供应商）-- 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 (≤105), 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:

Ki ID[1] ID[2] … ID[Ki]

where in the i-th line, Ki 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. Kj 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

解题思路

每个节点的卖出价格都是父节点价格的 $(1+r\%)$ 倍，消费者只会从叶子节点买入，价格是 $P*(1+r\%)^h$ , $h$ 是该叶子节点的高度，这里 $P, r$ 都是常量，所以只要找到最矮的叶子节点所在的高度，以及最矮叶子节点的个数(题目的另一个输出)
代码上只要在层序遍历的框架中套入寻找叶子节点的逻辑即可，层序遍历到的第一个叶子节点是最矮的，接着再记录该层叶子节点的数量

代码

#include <iostream>
#include <cstring>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
using namespace std;
#define DEBUG
#ifdef DEBUG
#define cin ifile
ifstream ifile("input.txt");
#endifint main(){int N;double P,r;cin>>N>>P>>r;vector< vector<int> > tree(N);int ki = 0,id = 0;for(int i=0;i<N;++i){cin>>ki;for(int j=0;j<ki;++j){cin>>id;tree[i].push_back(id);}}queue<int> q;q.push(0);//cur: index of current node//n_sup: sum of current node's child//level: the height of the current node//remain: remaining nodes of this height//n_cand: the amount of lowest leaf nodeint cur = 0,n_sup = 0,level=0,remain=1,n_cand = 0;while(!q.empty()){cur = q.front();q.pop();remain--;n_sup = tree[cur].size();//find leaf nodeif(n_sup == 0){n_cand=1;int candiate = 0;//find other leaf node in the same heightfor(int i=0;i<remain;++i){candiate = q.front();q.pop();if(tree[candiate].empty()){n_cand++;}}break;}for(int i=0;i<n_sup;++i){q.push(tree[cur][i]);}if(remain==0){level++;remain = q.size();}}printf("%.04f %d",P*pow(1+r/100,level),n_cand);return 0;
}