(PAT 1079) Total Sales of Supply Chain (树的深度优先遍历)

喜欢ヅ旅行 2022-03-21 05:12 149阅读 0赞

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 Nlines 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

解题思路:

先根据输入把树构建起来,K为0的为叶子节点,然后深度遍历树,叶子结点为retailer,叶子节点的权值\*(1+r/100)^深度\*原价就是,这位retailer售出的价格,把所有叶子结点的价格汇总就是总收益

#include <iostream>
    #include <algorithm>
    #include <vector>
    #include <math.h>
    using namespace std;
    int N;
    double ans = 0.0;
    double P, r;
    struct CNode {
    	bool isroot = true;
    	vector<int> child;
    }CTree[100010];
    int Retailers[100010];
    
    void DFS_C(int root, int depth) {
    	if (CTree[root].child.empty()) {
    		ans += P * pow((1 + r / 100), depth)*Retailers[root];
    	}
    	else {
    		for (int i = 0; i < CTree[root].child.size(); ++i) {
    			DFS_C(CTree[root].child[i], depth + 1);
    		}
    	}
    }
    
    int main() {
    	int K;
    	scanf("%d %lf %lf", &N, &P, &r);
    	for (int i = 0; i < N; ++i) {
    		scanf("%d", &K);
    		if (K == 0) {
    			int price;
    			scanf("%d", &price);
    			Retailers[i] = price;
    		}
    		else {
    			for (int j = 0; j < K; ++j) {
    				int node;
    				scanf("%d", &node);
    				CTree[i].child.push_back(node);
    				CTree[node].isroot = false;
    			}
    		}
    	}
    	
    	//寻找根
    	int root = -1;
    	for (int i = 0; i < N; ++i) {
    		if (CTree[i].isroot == true) {
    			root = i;
    			break;
    		}
    	}
    
    	DFS_C(root, 0);
    	printf("%.1lf", ans);
    
    	system("PAUSE");
    	return 0;
    }