PAT 1079. Total Sales of Supply Chain (25)（bfs，树，树的层数）

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1079. Total Sales of Supply Chain (25)

时间限制
250 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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

题目大意

• 1.p进货，获利r%，只有零售店面对消费者，假设除了根供应商，其他都只有一个供应商。
• 2.Ki ID[1] ID[2] … ID[Ki]，如果ki=0，代表它只是一个零售商，不供货直接面对消费者，后面跟着他的是其总产品的数量。如果ki！=0则后面跟着的是ki个买它货的供应商或者零售商。
• 3.输出零售商的总销售额，精确到小数点后一位数。

解题思路

• 1.题目就是一颗树，求出叶节点属于第几层，然后带入公式算出结果就行了。

AC代码

#include<iostream>#include<vector>#include<deque>#include<math.h>#include<algorithm>#include<cstdio>using namespace std;int n;double p, r;int isLoof[100001],lever[100001];vector<int> childrens[100001];int main(){    cin >> n >> p >> r;    int k,child;    for (int i = 0; i < n; i++)    {        cin >> k;        if (k!=0)        {            //代表不是叶节点，输入孩子            while (k--)            {                cin >> child;                childrens[i].push_back(child);            }        }        else        {            //如果k为零，代表零售店，则child为零售店的销售量            cin >> child;            isLoof[i] = child;        }    }    //bfs遍历并计算结果    double sum = 0;    deque<int> dq;    dq.push_back(0);    int now;    while (!dq.empty())    {        now = dq.front();        dq.pop_front();        if (!childrens[now].empty())        {            ;            for (int i = 0; i < childrens[now].size(); i++)            {                child = childrens[now][i];                lever[child] = lever[now] + 1;                dq.push_back(child);            }        }        else if (isLoof[now]!=0)        {            //如果没有孩子，则他是叶节点了，计算和            sum += isLoof[now] * p * pow((1 + r / 100.0), lever[now]);        }    }    printf("%.1f\n", sum);    return 0;}