1079 Total Sales of Supply Chain (25分)

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本文介绍了一个供应链模型，其中包含零售商、分销商和供应商，详细解释了产品从供应商到客户的价格递增过程。通过输入供应链成员数量、初始价格和价格增长率，文章提供了一种算法来计算所有零售商的总销售额。

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 total sales from all the 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 unit 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, then instead the total amount of the product will be given after Kj. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the total sales we can expect from all the retailers, accurate up to 1 decimal place. It is guaranteed that the number will not exceed 1010.

Sample Input:

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

Sample Output:

42.4

这个题是我第三遍做了，1106，不解释，直接莽！

#include<iostream>
#include<cmath>
#include<vector>
int a[100001] = {0};
int h_record[100001] = {0};
int height(int num){
    int h = 0;
    int temp_num = a[num];
    while (num != 0){ //竟然超时，哼哼，小骚一下～

        if(h_record[num]){
            return h + h_record[num];
        }
        h++;
        num = a[num];
    }
    h_record[temp_num] = h - 1;
    return h;
}
using namespace std;

int main(){
    int n;
    double p, r;
    cin >> n >> p >> r;
    
    vector<int> vec;
    vector<int> vec_temp;
    for(int i = 0; i < n; i++){

        int k;
        scanf("%d", &k);
        if(k){
            for(int j = 0; j < k; j++){
                int temp;
                scanf("%d", &temp);
                a[temp] = i;
            }
        }else
        {
            int temp;
            scanf("%d", &temp);
            vec.push_back(temp);
            vec_temp.push_back(i);
        }


    }
    double sum = 0;
    for(int i = 0; i < vec.size(); i++){

        sum += p * pow(1.0 + r / 100.0, height(vec_temp[i])) * vec[i];
    }

    printf("%.1f\n", sum);


    return 0;
}