Pat(A) 1106. Lowest Price in Supply Chain (25)

本文介绍了一种通过广度优先搜索算法(BFS)解决供应链中最低零售价格问题的方法。在一个由零售商、分销商和供应商组成的层级网络中,从根供应商开始,每层的价格会增加固定百分比。文章详细解释了如何利用BFS找到达到零售商的最短路径,并计算最低零售价格及其出现的次数。

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原题目:

原题链接:https://www.patest.cn/contests/pat-a-practise/1106

1106. Lowest Price in Supply Chain (25)


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 1010.

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

题目大意

有一系列经销商,除了根经销商,每个经销商有且只有一个上级,从上级拿到货之后会比进货价多r%卖出。问零售商中(没有下级)中最低价会是多少,有几个卖最低价的零售商。
输入数据为经销商数N(经销商标记为0~N-1),最原始价格,比例r
后N行为第i个经销商的下级数以及其下级。。

解题报告

本题与Pat甲级的1090题:1090. Highest Price in Supply Chain (25)有相似之处。
1090求树的深度,并求出该层有多少节点,求深度用的dfs。
本题求的是层次最小的叶子节点的层数,并求出该层多少叶子节点,用的是bfs。

代码

#include "iostream"
#include "vector"
#include "queue"
#include "cmath"
using namespace std;

vector<vector<int>> G;

int N;
double P,r;
int ansLevel,ansNum;

void init(){
    cin>>N>>P>>r;
    G.resize(N);
    int i,j,k,x;
    for(i = 0; i < N; i++){
        cin>>k;
        for(j = 0; j < k; j++){
            cin>>x;
            G[i].push_back(x);
        }
    }
}

void bfs(){
    queue<int> que;
    bool flag = false;
    int cnt = 0,num = 1;
    int nextNum = 0;
    ansLevel = 1;
    ansNum = 0;
    que.push(0);
    int x,i;
    while(!que.empty()){
        x = que.front();
        que.pop();
        cnt ++;
        if(G[x].size() == 0){
            flag = true;
            ansNum ++;
        }
        for(i = 0; i < G[x].size(); i++){
            que.push(G[x][i]);
            nextNum ++;
        }
        if(cnt == num){
            if(flag)
                return;
            ansLevel ++;
            num = nextNum;
            nextNum = 0;
            cnt = 0;
        }
    }
}

int main(){
    init();
    bfs();
    printf("%.4f %d\n",P * pow(1+ r / 100,ansLevel - 1),ansNum);
    system("pause");
}
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